THE APOLLONIAN STRUCTURE OF INTEGER
SUPERHARMONIC MATRICES
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
Abstract. We prove that the set of quadratic growths attainable by integervalued superharmonic functions on the lattice Z2 has the structure of an Apollonian circle packing. This completely characterizes the PDE which determines
the continuum scaling limit of the Abelian sandpile on the lattice Z2 .
1. Introduction
1.1. Main results. This paper concerns the growth of integer-valued superharmonic functions on the lattice Z2 ; that is, functions g : Z2 → Z with the property
that the value at each point x is at least the average of the values at the four lattice
neighbors y ∼ x. In terms of the Laplacian operator, these are the functions g
satisfying:
X
∆g(x) :=
(g(y) − g(x)) ≤ 0
(1.1)
y∼x
2
for all x ∈ Z . Our goal is to understand when a given quadratic growth at infinity,
specified by a 2 × 2 real symmetric matrix A, can be attained by an integer-valued
superharmonic function. For technical reasons, it is convenient to replace 0 by 1 in
the inequality above. (It is straightforward to translate between the two versions
using the function f (x) = 21 x1 (x1 +1), which has ∆f ≡ 1.) So we seek to determine,
for each matrix A, whether there exists a function g : Z2 → Z such that
g(x) = 21 xt Ax + o(|x|2 )
and
∆g(x) ≤ 1
for all x ∈ Z2 .
(1.2)
When this holds, we say that the matrix A is integer superharmonic, and we call g
an integer superharmonic representative for A.
We will relate the set of integer superharmonic matrices to an Apollonian circle
packing. Recall that every triple of pairwise tangent general circles (circles or
lines) in the plane has exactly two Soddy general circles tangent to all three. An
Apollonian circle packing is a minimal collection of general circles containing a given
triple of pairwise-tangent general circles that is closed under the addition of Soddy
general circles. Let Bk (k ∈ Z) denote the Apollonian circle packing generated by
the vertical lines through the points (2k, 0) and (2k + 2, 0) in R2 together S
with the
circle of radius 1 centered at (2k+1, 0). The band packing is the union B = k∈Z Bk ,
which is a circle packing of the whole plane, plus the vertical lines.
To each circle C ∈ B with center (x1 , x2 ) ∈ R2 and radius r > 0, we associate
the matrix
1 r + x1
x2
AC =
.
(1.3)
x2
r − x1
2
Date: May 22, 2014.
1
2
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
Figure 1.1. One 2Z2 -period of the boundary of the set of integer
superharmonic matrices, as characterized by Theorem 1.1.
We use the semi-definite order on the space S2 of 2 × 2 real symmetric matrices,
which sets A ≤ B if and only if B − A is positive semidefinite. Our main result
relates integer superharmonic matrices to the band packing.
Theorem 1.1. A ∈ S2 is integer superharmonic if and only if A ≤ AC for some
circle C ∈ B.
This theorem implies that the boundary of the set of integer superharmonic
matrices looks like the surface displayed in Figure 1.1: it is a union of slope-1 cones
whose bases are the circles C ∈ B and whose peaks are the matrices AC . Here
we have identified S2 with R3 with coordinates (x1 , x2 , r), and B lies in the r = 0
plane. This embedding has the property that the intersection of the cone
{A ∈ S2 | A ≤ AC }
with the r = 0 plane is the closed disk bounded by C.
To prove Theorem 1.1 we will recursively construct an integer superharmonic
representative gC for each matrix AC (C ∈ B). Let
LC = {v ∈ Z2 | AC v ∈ Z2 }.
(1.4)
By the extended Descartes circle theorem of Lagarias, Mallows and Wilks [13], each
AC has rational entries, so LC is a full-rank sublattice of Z2 .
The following theorem, from which Theorem 1.1 will follow, encapsulates the
essential properties of the gC ’s we construct.
Theorem 1.2. For each circle C ∈ B, there exists an integer superharmonic representative gC for AC which satisfies the periodicity condition
gC (x + v) = gC (x) + xt AC v + gC (v)
(1.5)
for all v ∈ LC and x ∈ Z2 . Moreover, gC is maximal in the sense that g − gC is
bounded whenever g : Z2 → Z satisfies ∆g ≤ 1 and g ≥ gC .
INTEGER SUPERHARMONIC MATRICES
3
Figure 1.2. Periodic pattern ∆gC of the integer superharmonic
representative Theorem 1.2, shown for five different circles C.
Black, patterned, and white cells correspond to sites x ∈ Z2 where
∆gC (x) equals 1, 0, and −2, respectively. In each case the fundamental tile TC is identified by a white outline. Clockwise from
the top left is ∆gC for the circle (153, 17, 120), its three parents
(4, 1, 4), (9, 1, 6), and (76, 7, 60), and its Soddy precursor (25, 1, 20).
The circles themselves are drawn in Figure 1.3.
The condition (1.5) implies that gC (x) − 21 xt AC x − bt x is periodic for some b ∈ R2
and we call an integer superharmonic representative with this property an odometer
for AC . The construction of gC is explicit but rather elaborate. In Section 2 we
4
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
C2
C4
C1
C0
C3
Figure 1.3. The child circle C0 = (153, 17, 120) and its parents
C1 = (76, 7, 60), C2 = (4, 1, 4), and C3 = (9, 1, 6) in B. The other
Soddy circle C4 = (25, 1, 20) ∈ B for the parents of C0 is the Soddy
precursor of C0 . The triples of integers are curvature coordinates,
defined in Section 1.3.
outline the steps of this construction and give the derivation of Theorem 1.1 from
Theorem 1.2. We now briefly survey a few connections to our work.
1.2. Hexagonal tilings of the plane by 90◦ symmetric tiles. Figure 1.2 shows
the Laplacians ∆gC for a triple of circles in B and their two Soddy circles. In each
case ∆gC is periodic and we have outlined a fundamental domain TC on whose
boundary ∆gC = 1. A major component of our paper is the construction of these
TC , which turn out to have a remarkable tiling property.
To state it precisely, for x ∈ Z2 write sx = {x1 , x1 + 1} × {x2 , x2 + 1} ⊆ Z2 and
s̄x = [x1 , x1 + 1] × [x2 , x2 + 1] ⊆ R2 . If T is a set of squares sx , we call T a tile if
the set
[
I(T ) :=
s̄x ⊆ R2
(1.6)
sx ∈T
is a topological disk. A tiling of Z2 is a collection of tiles T such that every square
sx (x ∈ Z2 ) belongs to exactly one tile from T .
Theorem 1.3. For every circle C ∈ B, there is a tile TC ⊆ Z2 with 90◦ rotational
symmetry, such that TC + LC is a tiling of Z2 . Moreover, except when C has radius
1, each tile in TC + LC borders exactly 6 other tiles.
The tiles TC have the peculiar feature of being more symmetric than their plane
tilings (which are only 180◦ symmetric). We expect this to be a strong restriction.
In particular, call a tiling regular if it has the form T + L for some tile T and lattice
L ⊆ Z2 , and hexagonal if each tile borders exactly 6 other tiles. For regular tilings
T , T 0 of Z2 , write T 0 ≺ T if each tile in T ∈ T is a union of tiles from T 0 , and call
the regular tiling T primitive if T 0 ≺ T implies that either T 0 = T , or T 0 is the
tiling of Z2 by squares sx .
Conjecture 1.4. If T is a primitive, regular, hexagonal tiling of Z2 by 90◦ symmetric tiles, then T = TC + LC + v for some C ∈ B and some v ∈ Z2 .
INTEGER SUPERHARMONIC MATRICES
5
Figure 1.4. The tile of the circle (153, 17, 120), decomposed into
the tiles of its parents and Soddy precursor.
1.3. Apollonian circle packings. Our proof of Theorem 1.2 is a recursive construction which mimics the recursive structure of Apollonian circle packings. Identifying a circle with center (x1 , x2 ) ∈ R2 and radius 1c ∈ R with its curvature
coordinates (c, cx1 , cx2 ) (some care must be taken in the case of lines), the Soddy
circles C0 and C4 of a pairwise tangent triple of circles C1 , C2 , C3 satisfy the linear
equality
C0 + C4 = 2(C1 + C2 + C3 ).
(1.7)
This is a consequence of the Extended Descartes Theorem of Lagarias, Mallows,
and Wilks [13], and can be used, for example, to prove that every circle in B has
integer curvature coordinates. Pairwise tangent circles C1 , C2 , C3 , C4 constitute a
Descartes quadruple. Under permutation of indices, (1.7) gives four different ways
of producing a new Descartes quadruple sharing three circles in common with the
original. These four transformations correspond to the four generators of the Apollonian group of Graham, Lagarias, Mallows, Wilks, and Yan [11], which acts on the
Descartes quadruples of a circle packing. Our proof works by explicitly determining the action of the same Apollonian group on the set of maximal superharmonic
representatives, by giving an operation on our family of integer superharmonic representatives analogous to the operation (1.7) for circles. In particular, referring to
Figure 1.4, the example tile T0 is seen to decompose into 2 copies each of T1 , T2 , T3 ,
with the copies of T1 overlapping on a copy of T4 ; note, for example, that the tile
areas must therefore satisfy |T0 | + |T4 | = 2(|T1 | + |T2 | + |T3 |).
It follows from (1.7) that if the three generating general circles of an Apollonian
circle packing have integer curvatures, then every general circle in the packing
has integer curvature. In such a packing, the question of which integers arise as
curvatures has attracted intense interest over the last decade [2, 3, 11, 12, 22]; see
[10] for a survey.
Theorem 1.1 can be regarded as a new characterization of the circles appearing
in the band packing: the circles in B correspond to integer superharmonic matrices
that are maximal in the semidefinite order. In constructing gC and TC by an
analogue of the Descartes rule (1.7), we follow ideas of Stange [23], who associates
circles to “lax lattices” and proves a Descartes rule relating the bases of lattices
corresponding to four mutually tangent general circles. Our proof also associates
to each circle in B more detailed arithmetic information: Theorems 1.2 and 1.3
associate to each circle C ∈ B an integer superharmonic representative gC : Z2 → Z
and a fundamental tile TC . The curvature of C can be recovered as the area of TC .
1.4. Abelian sandpile. We briefly describe the Abelian sandpile model of Bak,
Tang, and Wiesenfeld [1] that motivated our work. Put n chips at the origin of Z2 .
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LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
In the sandpile model, a vertex having at least 4 chips topples by sending one chip
to each lattice neighbor. Dhar [6] observed that the final configuration of chips
does not depend on the order of topplings. This final configuration sn displays
impressive large-scale patterns [4, 7, 16, 21] and has been proposed by Dhar and
Sadhu as a model of “proportionate growth” [8]: as n increases, patterns inside
sn scale up in proportion to the size of the whole. The second and third authors
[19] have shown that as n → ∞ the sandpile sn has a scaling limit on Rd . The
sandpile PDE that describes this limit depends on the set of integer superharmonic
d × d matrices. In the companion paper [14] we use Theorem 1.1 to construct exact
solutions to the sandpile PDE. An intriguing open problem is to describe the set of
integer superharmonic matrices and analyze the associated sandpile processes for
periodically embedded graphs other than Z2 . (See [18].)
The weak-∗ limiting sandpile s : R2 → R appears to have the curious property that it is locally constant on an open neighborhood of the origin. Regions of
constancy in s correspond to regions of periodicity in sn . Ostojic [16] proposed
classifying which periodic patterns occur in sn . Caracciolo, Paoletti, and Sportiello
[4] and Paoletti [17] give an experimental protocol that recursively generates 2dimensional periodic “backgrounds” and 1-dimensional periodic “strings.” While
this protocol makes no explicit reference to Apollonian circle packings, we believe
that the 2-dimensional backgrounds it generates are precisely the Laplacians ∆gC
for C ∈ B. Moreover, periodic regions in sn empirically correspond to the Laplacians ∆gC of our odometers gC (C ∈ B), up to error sets of asymptotically negligible
size,
Acknowledgments. We appreciate several useful discussions with Guglielmo Paoletti, Scott Sheffield, Andrea Sportiello, Katherine Stange, András Szenes, and
David Wilson. The authors were partially supported by NSF grants DMS-1004696,
DMS-1004595, and DMS-1243606, respectively.
2. Overview of the proof
Here we outline the main elements of the proof of Theorem 1.2, and derive
Theorem 1.1 from Theorem 1.2. The example in Figure 1.4 motivates the following
recursive strategy for constructing the odometers gC :
(1) Construct fundamental domains (tiles) TC for each circle C ∈ B such that
(a) TC tiles the plane periodically; and
(b) TC decomposes into copies of smaller tiles TC 0 with specified overlaps.
(2) Use the decomposition of TC to recursively define an odometer on TC .
(3) Extend the odometer to Z2 via the periodicity condition (1.5), and check
that the resulting extension gC satisfies ∆gC ≤ 1 and is maximal in the
sense stated in Theorem 1.2.
In Section 3, we carry out this strategy completely for two especially simple classes
of circles.
For the general case of Step 1, we begin by associating to each pair of tangent
circles C, C 0 ∈ B a pair of Gaussian integers v(C, C 0 ), a(C, C 0 ) ∈ Z[i], in Section 4.
The v(C, C 0 )’s will generate our tiling lattices, while the a(C, C 0 )’s will describe
affine relationships among tile odometers. Recursive relations for these v’s and a’s,
collected in Lemma 4.1, are used extensively throughout the paper.
INTEGER SUPERHARMONIC MATRICES
7
In Section 6 we construct the tile TC recursively by gluing copies of tiles of
the parent circles of C as pictured in Figure 1.4, using the vectors v(C, C 0 ) to
specify the relative locations of the subtiles. The difficulty now lies in checking
that the resulting set is in fact a topological disk and tiles the plane. This is proved
in Lemma 6.3, which recursively establishes certain compatibility relations for tiles
corresponding to tangent circles in B. The argument uses a technical lemma, proved
in Section 5, which allows one to prove that a collection of tiles forms a tiling from
its local intersection properties.
Section 8 carries out Step 2 by constructing a tile odometer (an integer-valued
function on TC ) for each C ∈ B. The recursive construction of the tile odometers
mirrors that of the tiles (compare Definitions 6.1 and 8.3) using the additional data
of the Gaussian integers a(C, C 0 ), but we must check that it is well-defined where
subtiles intersect. A crucial difference between the tiles and the tile odometers
is that the former have 90◦ rotational symmetry, a fact which is exploited in the
proof of the key Lemma 6.3. As tile odometers are merely 180◦ symmetric, we
require an extra ingredient to assert compatibility of some pairs of tile odometers.
In particular, we show in Section 7 that tile boundaries can be suitably decomposed
into smaller tiles, a fact which is used in the induction to verify some of compatible
tile odometer pairs.
The final step is carried out in Section 9. In Lemma 9.1 we use the compatibility
relations among tile odometers to show that each tile odometer has a well-defined
extension gC : Z2 → Z satisfying (1.5). We then show ∆gC ≤ 1 by giving an explicit
description of the values of the Laplacian ∆gC on subtile intersections (Lemma 9.3).
In Lemma 9.4, we use this same explicit description of ∆gC to prove maximality
(the fact that ∆gC ≡ 1 on the boundary of TC will be crucial here). The proof of
Theorem 1.2 is completed by showing that the lattice LC of (1.4) is the same as the
lattice ΛC generated by the vectors v(C, Ci ), where C1 , C2 , C3 are the parent circles
of C. (Amusingly, our proof of this elementary statement uses the construction of
the odometers gC , although we suspect that a simpler proof could be found.)
Proof of Theorem 1.1. We conclude this section by deriving Theorem 1.1 from Theorem 1.2. The main observation is that the set of integer superharmonic matrices
is downward closed in the semidefinite order: if A ≤ B and B is integer superharmonic, then A is integer superharmonic. The following lemma, whose proof is
essentially just a double Legendre transform, demonstrates this fact by giving a
suitable integer approximation of the difference x 7→ 12 xt (B − A)x.
Lemma 2.1. If A ∈ S2 satisfies A ≥ 0, then there exists g : Z2 → Z such that
∆g(x) ≥ 0
and
g(x) = 21 xt Ax + O(1 + |x|)
for all x ∈ Z2 .
Proof. Define q(x) = 12 xt Ax and
g(x) = sup inf2 bq(y) + |y|c + p · (x − y).
p∈Z2 y∈Z
8
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
If we set p = bDq(x)c in the supremum, then we obtain
g(x) ≥ inf2 q(y) + |y| + bDq(x)c · (x − y) − 1
y∈Z
≥ inf2 q(y) + Dq(x) · (x − y) − |x| − 1
y∈Z
≥ q(x) − |x| − 1,
where we used the triangle inequality in the second step and the convexity of q in
the third. If we set y = x in the infimum, then we obtain
g(x) ≤ sup q(x) + |x| + 1 = q(x) + |x| + 1.
p∈Z2
Finally, since g is a pointwise supremum of affine functions, we obtain ∆g ≥ 0 by
the monotonicity of ∆.
The downward closure of integer superharmonic matrices now makes the “if”
direction of Theorem 1.1 an immediate consequence of Theorem 1.2.
For the “only if” direction, observe that points of tangency are dense on each
circle of the band packing B. In particular, any circle in the plane is either enclosed
by some circle of B or strictly encloses some circle of B. Therefore any matrix
A ∈ S2 satisfies either A ≤ AC for some C ∈ B or AC < A (that is, A − AC is
positive definite) for some C ∈ B. In the latter case, the existence of an integer
superharmonic representative g for A would contradict the maximality of gC : by
(1.1), some constant offset g + c of g would satisfy g + c ≥ gC , but AC < A implies
that g + c − gC is unbounded.
3. Two Degenerate Cases
3.1. Explicit constructions. Before defining odometers for general circles, we
consider two subfamilies of B, the Ford and Diamond circles. The odometers for
these subfamilies have simpler structure than the general case, and this allows us to
give a more explicit description of their construction. The purpose of carrying this
out is twofold. First, these cases, particularly the Ford circles, serve as a concrete
illustration of our general construction that avoids most of the technical complexity.
Second, by taking these subfamilies as our base-case in the general construction,
we avoid several tedious degeneracies. In particular, the general construction operates quite smoothly when the circle under consideration has parents with distinct
non-zero curvatures; this condition fails precisely when the circle is either Ford or
Diamond.
3.2. The Ford circles. It is a classical fact that the reduced fractions, namely,
the pairs of integers (p, q) ∈ Z2 such that q ≥ 1 and gcd(q, p) = 1, are in algebraic
bijection with the circles in B0 that are tangent to the vertical line through the
origin. In curvature coordinates, this bijection is given by
(p, q) 7→ (q 2 , 1, 2pq)
and the circles are called the Ford circles. We write Cpq for the Ford circle associated
to the reduced fraction (p, q).
The tangency structure of the Ford circles is famously simple. If q ≥ 2, then the
parents of the Ford circle Cpq in the packing B0 (Figure 4.1) are the vertical line
1
2
through the origin, and the unique Ford parents Cpq
:= Cp1 q1 and Cpq
:= Cp2 q2
determined by the constraints p1 q − q1 p = 1, p2 q − q2 p = −1, 0 ≤ q1 < q and
INTEGER SUPERHARMONIC MATRICES
9
C2,5
C3,8
C1,3
Figure 3.1. The Ford circles C1,3 , C2,5 , and C3,8 and several
periods of the Laplacian of their odometers gpq : Z2 → Z.
E4
E2
E4
E2
E4
E2
E1
E3
E1
E3
E1
E3
Figure 3.2. The three possible overlapping structures of the subtiles Ei of a Ford tile [0, q]2 .
0 < q2 ≤ q. The Ford parents are the only Ford circles tangent to Cpq having
smaller curvature (except that when q = 1, we have q1 = 0 and thus the Ford
parents are not both Ford circles). We observe that the definition of parents implies
(p, q) = (p1 , q1 ) + (p2 , q2 ) and p1 /q1 > p/q > p2 /q2 , which gives the classical
relationship between the Ford circles and the Farey fractions. This connection
between the circles and Farey fractions was used by Ford [9] (and later, by Nichols
[15]) to prove results about Diophantine approximation.
The odometer gpq : Z2 → Z associated to the Ford circle Cpq also enjoys a
simple description. To understand the structure of gpq , we consider the Ford circle
C3,8 and its parents C2,5 and C1,3 . These circles have the tangency structure and
periodic odometer Laplacians ∆gpq displayed in Figure 3.1.
Examining the patterns in Figure 3.1, we see that [1, q]2 is a fundamental domain for the periodic Laplacian ∆gpq . Moreover, the fundamental tile of ∆g3,8
decomposes (with a few errors on points of overlap) into two copies each of the
fundamental tiles of ∆g1,3 and ∆g2,5 .
These observations lead to the following construction. For a general Ford circle
Cpq , we define an odometer gpq : Z2 → Z by specifying its values on the domain [0, q]2 ⊆ Z2 and then extending periodically. The values of gpq on [0, q]2
are determined recursively by copying data from the parent odometers gp1 q1 and
gp2 q2 onto the subdomains E1 := [0, q1 ]2 , E2 := [q2 , q]2 , E3 := [q1 , q] × [0, q2 ], and
E4 := [0, q2 ] × [q1 , q]. These subdomains always have one of the three overlapping
structures displayed in Figure 3.2, depending on the relative sizes of q1 and q2 . This
construction is encoded precisely in the following lemma.
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LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
Lemma 3.1. There is a unique family of functions gpq : Z2 → Z, indexed by the
Ford circles Cpq , satisfying
gpq (x) = d pq x1 x2 e
for x ∈ [0, q]2 \ [2, q − 2]2
gpq (x + (0, −q)) = gpq (x) + (−p, 0) · x
for x ∈ Z2
gpq (x + (q, q1 )) = gpq (x) + (p1 , p) · x + q1 p
for x ∈ Z2
gpq (x + (−q, q2 )) = gpq (x) + (p2 , −p) · x − qp2
for x ∈ Z2
and, when q ≥ 2,
gpq (x) = gp1 q1 (x)
for x ∈ E1
gpq (x) = gp1 q1 (x − (q2 , q2 )) + (p2 , p2 ) · x − p2 q2 + 1
for x ∈ E2
gpq (x) = gp2 q2 (x − (q1 , 0)) + (0, p1 ) · x
for x ∈ E3
gpq (x) = gp2 q2 (x − (0, q1 )) + (p1 , 0) · x
for x ∈ E4 ,
where E1 := [0, q1 ]2 , E2 := [q2 , q]2 , E3 := [q1 , q] × [0, q2 ], and E4 := [0, q2 ] × [q1 , q].
Proof. We prove by induction on q ≥ 1 that there is a function gpq : Z2 → Z
satisfying the above conditions. For p ∈ Z, we observe that
gp,1 (x) := 21 x1 (x1 − 1) + px1 x2
satisfies the first four rules. Thus, we may assume that q ≥ 2 and the induction
hypothesis holds when q 0 < q.
We first observe that any map gpq : [0, q]2 → Z that satisfies the first condition
has a unique extension to all of Z2 that satisfies the next three conditions. Since
the translations of [0, q]2 by the lattice Lpq generated by {(0, −q), (q, q1 ), (−q, q2 )}
cover Z2 , the three translation conditions prescribe values for gpq on the rest of
Z2 . Since [0, q − 1]2 + Lpq is a partition of Z2 and the composition of the three
translations is the identity, it suffices to check the consistency of the translation
conditions on the set [0, q]2 \ [1, q − 1]2 . This follows from the first condition.
Thus, to construct gpq , it suffices to check the consistency of the first and the
last four conditions, which specify the values of gpq on [0, q]2 . We observe that the
first condition prescribes values on the doubled boundary set B := [0, q]2 \ [2, q − 2]2
and the last four conditions prescribe values on the sets Ei described above. We
must check that the consistency of these five prescriptions.
Case 1: consistency for the intersections E1 ∩ E3 , E1 ∩ E4 , E2 ∩ E3 , and E2 ∩ E4 .
To check the first intersection, we must verify that
gp1 q1 (x) = gp2 q2 (x − (q1 , 0)) + (0, p1 ) · x,
for x ∈ E1 ∩ E3 = {q1 } ∩ [0, min{q1 , q2 }]. Applying the inductive version of the first
condition, this reduces to
p1
p2
x1 x2 = (x1 − q1 )x2 + p1 x2 .
q1
q2
Since x1 = q1 , this is easily seen to be true. The other three intersections can by
checked by symmetric arguments.
Case 2: consistency for the intersections E1 ∩ E2 and E3 ∩ E4 . To check the
first intersection, we must verify that
gp1 q1 (x) = gp1 q1 (x − (q2 , q2 )) + (p2 , p2 ) · x − p2 q2 + 1,
INTEGER SUPERHARMONIC MATRICES
11
for all x ∈ E1 ∩ E2 = [q2 , q1 ]2 . This is non-trivial if and only if q1 ≥ q2 , in
which case Cp2 q2 is a parent of Cp1 q1 . Let Cp3 q3 denote the other parent. Since
p3 /q3 > p1 /q1 > p2 /q2 , the inductive version of the last four conditions gives
gp1 q1 (x) = gp3 q3 (x − (q2 , q2 )) + (p2 , p2 ) · x − p2 q2 + 1
and, since q3 = q1 − q2 ,
gp1 q1 (x − (q2 , q2 )) = gp3 q3 (x − (q2 , q2 )),
2
for all x ∈ [q2 , q1 ] . Substituting this into the above equation, we easily see that
equality holds. The other intersection can be checked by a symmetric argument.
Case 3: consistency for the intersections B ∩ Ei . When i = 1, this amounts to
verifying
gp1 q1 (x) = d pq x1 x2 e
for x ∈ [0, q1 ] × [0, 1]. By induction, this reduces to checking
d pq11 x1 x2 e = d pq x1 x2 e,
for x ∈ [0, q1 ] × [0, 1]. When x1 x2 = 0, this is trivial, so we may assume x1 > 0 and
x2 = 1. Since gcd(p, q) = 1, we see that the distance between pq x1 and the nearest
integer is at least 1q . Using the relation qp1 − pq1 = 1, we see that | pq x1 − pq11 x1 | =
p
p1
1
1
qq1 |x1 | ≤ q and therefore d q1 x1 x2 e = d q x1 x2 e, as desired. The other intersections
can be checked by symmetric arguments.
In contrast to our construction of general circle odometers in later sections,
the description of the Ford circle odometers gpq in Lemma 3.1 has two enormous
advantages: the fundamental tiles are square shaped and, more importantly, there
is a closed formula for the odometer on the outer two layers of the fundamental
tile. This renders geometric questions about the fundamental tile moot and makes
it possible to compute the Laplacian ∆gpq in a relatively straightforward manner.
Define the boundary of a subset X ⊆ Z2 to be
∂X = {x ∈ X : |x − y| < 2 for some y ∈ Z2 \ X},
and call X \ ∂X the interior of X. In the case of a square [0, q]2 , we have ∂[0, q]2 =
[0, q]2 \ [1, q − 1]2 .
Lemma 3.2. The Laplacian ∆gpq : Z2 → Z satisfies
∆gpq (x) = ∆gpq (x + (0, q)) = ∆gpq (x + (q, q1 ))
for x ∈ Z2
and
∆gpq (x) = 1
for x ∈ [0, q]2 \ [1, q − 1]2 .
Moreover, if q ≥ 2 and x ∈ [1, q − 1]2 lies in 2, 3, or 4 of the boundaries of the sets
Ei , then ∆gpq (x) = 1, 0, or −2, respectively.
Proof. To obtain the periodicity conditions, we simply compute the Laplacian of
the periodicity conditions in Lemma 3.1, recalling that the Laplacian of an affine
function vanishes.
We next check ∆gpq (x) = 1 for x ∈ [0, q]2 \ [1, q − 1]2 . We claim that, for
0 ≤ a ≤ q,
gpq (a, 0) = gpq (0, a) = 0,
gpq (a, 1) = gpq (1, a) = d pq ae,
12
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
and
gpq (a, −1) = gpq (−1, a) = −b pq ac.
The first two are immediate from Lemma 3.1, so we check the third. Using the
periodicity, we compute
gpq (a, −1) = gpq (a, q − 1) − (p, 0) · (a, −1) = d pq a(q − 1)e − pa = −b pq ac.
Similarly, we compute
gpq (−1, a) = gpq (q − 1, a + q1 ) − (p1 , p) · (−1, a) − q1 p
= d pq (q − 1)(a + q1 )e + p1 − pa − q1 p
= −b pq (a + q1 )c + p1
= −b pq a − 1q c
= −b pq ac,
for 0 ≤ a ≤ q2 and
gpq (−1, a) = gpq (q − 1, a − q2 ) − (−p2 , p) · (−1, a) + p2 q
= d pq (q − 1)(q − q2 )e + p2 − pa + p2 q
= −b pq (a − q2 )c + p2
= −b pq a − 1q c
= −b pq ac,
for q2 ≤ a ≤ q. Using gcd(p, q) = 1, we then obtain
∆gpq (0, a) = ∆gpq (a, 0) = d pq ae − b pq ac = 1
for 0 ≤ a < q. That ∆gpq = 1 on the rest of [0, q]2 \ [1, q − 1]2 follows by periodicity.
The moreover clause can be handled similarly, using Lemma 3.1 to explicitly
evaluate the Laplacian at the intersections of the boundaries of the sets. For example, suppose that x ∈ (E1 ∩ E3 ) \ (E2 ∪ E4 ). In this case, x = (q1 , h) for some
0 < h < q1 , q2 . Using gpq = gp1 q1 on E1 , we compute
gpq (x + (0, k)) = d pq11 q1 (h + k)e = p1 (h + k)
for k = −1, 0, 1,
and
gpq (x − (1, 0)) = d pq11 (q1 − 1)he = p1 h − b pq11 hc.
Using qpq (y) = gp2 q2 (y − (q1 , 0)) + (0, p1 ) · x for y ∈ E3 , we compute
gpq (x + (1, 0)) = gp2 q2 (1, h) + p1 h
= d pq22 he + p1 h
= d pq11 h −
=
d pq11 he
1
q1 q2 he
+ p1 h
+ p1 h.
Putting these together, we obtain ∆gpq (x) = 1.
The above lemma tells us how to compute ∆gpq . Indeed, suppose we wish
to compute ∆gpq (x) for some x ∈ Z2 . We first reduce to the case x ∈ [0, q]2
using the periodicity of ∆gpq . Now, if x lies on the boundary of [0, q]2 or least
two of the boundaries of the Ei , then we can read off ∆gpq (x) from Lemma 3.2.
Otherwise, x lies in the interior of one of the Ei , and we can proceed recursively to
the corresponding parent ∆gpi qi .
INTEGER SUPERHARMONIC MATRICES
13
Following this line of reasoning, we prove Theorem 1.2 in the special case of the
Ford circles. Observe that the peak associated to the Ford circle (p, q) is the matrix
1 1 pq
Apq := 2
,
q pq 0
which has lattice Lpq := {x ∈ Z2 : Apq x ∈ Z2 } generated by {(0, −q), (q1 , q)}.
Proposition 3.3. For every Ford circle Cpq , the odometer gpq : Z2 → Z satisfies
∆gpq (x) ≤ 1
and
gpq (x + v) = gpq (x) + x · Apq v + gpq (v),
for all x ∈ Z2 and v ∈ Lpq . Moreover, the only infinite connected subset X ⊆ Z2
such that ∆(gpq + 1X ) ≤ 1 is X = Z2 .
Proof. The periodicity condition is immediate from Lemma 3.1, after we observe
Apq (0, q) = (p, 0), Apq (q, q1 ) = (p1 , p), gpq (0, q) = 0, and gpq (q, q1 ) = q1 p. That
∆gpq ≤ 1 is immediate from Lemma 3.2.
To check the moreover clause, we suppose X ⊆ Z2 is infinite and connected and
let X C = Z2 \ X denote its complement. If X̃ is the complement of any connected
component of X C , then ∆1X̃ ≤ ∆1X . In particular, we may assume that both X
and its complement X C are connected.
If X C intersects an Lpq translation of the boundary of [0, q]2 , then, since X is
infinite and connected, the set {x ∈ X C : ∆1X (x) > 0} must intersect an Lpq
translation of the boundary of [0, q]2 . By Lemma 3.2, ∆gpq (x) = 1 at any such
point. Thus, we may assume that X C is contained in the interior of [0, q]2 . In
particular, it suffices to prove the following claim.
Claim. Suppose Y ⊆ [0, q]2 is simply connected, Y ∩ [1, q − 1]2 is non-empty,
Y \ [1, q − 1]2 is connected, and (Y \ [1, q − 1]2 ) ∩ Ei is non-empty for at most one
Ei . Then there is a point x ∈ Y ∩ [1, q − 1]2 such that ∆gpq (x) − ∆1Y (x) > 1.
We proceed by induction on q. This is trivial when q = 1, since [1, q − 1]2 is
empty. When q = 2, we must have Y = {(1, 1)} ⊆ [0, 2]2 and, by Lemma 3.2, we
must have ∆gpq (1, 1) = −2. We may therefore assume q > 2. By symmetry, we
may assume q1 > q2 , so the Ei enjoy the intersection structure displayed on the left
of Figure 3.2. We divide the analysis into several cases.
Case 1. Suppose Y is contained in the interior of some Ei . We apply the
induction hypothesis to find x.
Case 2. Suppose x ∈ ∂Y ∩ [1, q − 1]2 lies in the boundary of exactly two Ei .
Since Lemma 3.2 gives ∆gpq (x) = 1, we have ∆(gpq − 1Y )(x) > 1.
Case 3. Suppose x ∈ ∂Y ∩ [1, q − 1]2 lies in the boundary of exactly three Ei .
Since we are not in the previous case, there are least two neighbors not in Y and
thus ∆1Y (x) < −2. By Lemma 3.2, ∆gpq (x) = 0.
Case 4. Suppose none of the above cases hold. Assuming q1 > q2 , then Y does
not intersect E3 or E4 . Moreover, it must not intersect either E1 \ [1, q − 1]2 or
E1 \ [1, q − 1]2 . By symmetry, we may assume that former. If Y ⊆ E1 , then we
can apply the induction hypothesis to Y . Otherwise, we can apply the induction
hypothesis to Y ∩ E2 . The case q2 > q1 is symmetric.
The proof of Theorem 1.2 for the Ford circles is completed by the following
lemma, which shows that the moreover clause of Proposition 3.3 suffices to show
maximality.
14
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
Lemma 3.4. If g : Z2 → Z is superharmonic, ∆g is doubly periodic, and the only
infinite connected set X ⊆ Z2 such that ∆(g + 1X ) ≤ 1 is X = Z2 , then g is
maximal in the sense of Theorem 1.2.
Proof. Suppose for contradiction that h : Z2 → Z satisfies h ≥ g and ∆h ≤ 1, and
that h − g is unbounded. Replacing h with h − minx∈Z2 (h − g), we may assume
that {h − g = 0} ⊆ Z2 is non-empty. Using the monotonicity of the Laplacian,
we see that f = min{h, g + 1} satisfies ∆f ≤ 1 and f − g ∈ {0, 1}. Since h − g
is unbounded and ∆(h − g) is bounded above (here we use the periodicity of ∆g),
the set {f − g = 1} ⊆ Z2 must contain connected components of arbitrarily large
size. Note that, if X is any such component, then we have ∆(g + 1X ) ≤ 1. Using
the periodicity of ∆g and compactness, we can select an infinite connected X ( Z2
such that ∆(g + 1X ) ≤ 1. Indeed, let Xn be a sequence of connected components of
{f −g = 1} such that |Xn | → ∞. Since {f −g = 0} ⊆ Z2 is non-empty, we may also
select xn ∈ {f − g = 0} which is adjacent to Xn . Using the fact that ∆g is periodic,
we can translate so that xn always lies in the same period of ∆g. Now, we can pass
to a subsequence nk such that xnk = x∗ is constant and Xm ∩{y ∈ Z2 : |y−x∗ | ≤ n}
is equal for all m ≥ n. Let X ∗ be the infinite connected component of limk→∞ Xnk
that is adjacent to x∗ . Then ∆g + ∆1X ∗ ≤ 1, contradicting our hypothesis.
3.3. The diamond circles. The diamond circles are the circles (c, cx1 , cx2 ) ∈ B0
which are tangent to (1, 1, 0) and (1, 1, 2) and satisfy 0 < x1 < 1. In curvature
coordinates, the diamond circles can be parameterized via
Ck := (2k(k + 1), 2k 2 − 1, 2k(k + 1)),
for k ∈ Z+ . The peak matrix associated to the diamond circle Ck is
k
1 k+1
1
Ak :=
1−k ,
2 1
k
which has lattice Lk := {x ∈ Z2 : Ak x ∈ Z2 } generated by {(0, −2k), (k + 1, k)}.
To better understand the structure of the diamond circle odometers gk : Z2 → Z,
we examine their Laplacians. Figure 3.3 displays the patterns associated to the first
four diamond circles. We observe that the periodicity of each pattern is exactly Lk .
Moreover, the internal structure of the fundamental tiles are similar enough that
we can immediately conjecture what the general case should be. In fact, for the
diamond circles, we have a closed formula for the odometer on each fundamental
tile, making this family even simpler than the Ford circles.
Lemma 3.5. For each k ∈ Z+ , there is a unique function gk : Z2 → Z such that
gk (x + (0, −2k)) = gk (x) + (−k, k − 1) · x − k(k − 2)
for x ∈ Z2 ,
gk (x + (k + 1, k)) = gk (x) + (k, 1) · x + 21 k(k + 1)
for x ∈ Z2 ,
gk (x + (−k − 1, k)) = gk (x) + (0, −k) · x − 12 k(k + 1)
for x ∈ Z2 ,
gk (x) = 12 |x1 |(|x1 | − 1) − b 14 (x1 − x2 )2 c
for x ∈ Tk ,
and
where
Tk := {x ∈ Z : max{|x1 |, |x2 − k|, |x1 | + |x2 − k| − 1} ≤ k}.
INTEGER SUPERHARMONIC MATRICES
15
C3
C1
C2 C4
Figure 3.3. Several periods of the Laplacian of the odometers
gk : Z2 → Z of the first four diamond circles. Here, the black, dark
patterned, light patterned, and white cells correspond to Laplacian
values 1, 0, −1, and −2, respectively.
Proof. This is similar to the corresponding proof for the Ford circles above. In
particular, it suffices to check the consistency of the translation conditions on the
set Tk \ Tk0 , where
Tk0 := {x ∈ Z : |x1 | + |x2 − k| ≤ k − 1}
is the interior of Tk . We claim that this follows from the fourth condition. If
x, x + (k + 1, k) ∈ Tk , then either x = (−j, j) for j = 1, ..., k − 1 or x = (−1 − j, j)
for j = 0, ..., k − 1. We then compute
gk (x + (k + 1, k)) = gk (x) + (k, 1) · x + 21 k(k + 1)
in either case. This implies the consistency of the second condition. The first and
third conditions follow similarly.
Lemma 3.6. For each k ∈ Z+ , the Laplacian ∆gk satisfies
∆gk (x + (0, 2k)) = ∆gk (x + (k + 1, k)) = ∆gk (x)
for x ∈ Z2
∆gk (x) = 1
for x ∈ Tk \ Tk0
∆gk (x) = (−1)x1 +x2 − 1{0} (x1 )
for x ∈ Tk0 .
Proof. The first two conditions follow by computations analogous to those in the
proof of Lemma 3.2. The formula in the third condition follows by inspection of
the formula for gk on Tk .
Proposition 3.7. For every diamond circle Ck , the odometer gk : Z2 → Z satisfies
∆gk (x) ≤ 1
and
gk (x + v) = gk (x) + x · Ak v + gk (v)
for all x ∈ Z2 and v ∈ Lk . Moreover, the only infinite connected subset X ⊆ Z2
such that ∆(gk + 1X ) ≤ 1 is X = Z2 .
Proof. This follows from the above lemma and the first part of the argument of
Proposition 3.3. We can avoid the recursive argument in this case because of the
explicit computation of ∆gk on Tk0 in Lemma 3.6.
16
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
4. Circles and Lattices
4.1. Periodicity conditions. When defining odometers for the Ford and diamond
circles above, we extended a finite amount of data to all of Z2 via periodicity
conditions of the form
g(x + v) = g(x) + a · x + k
for x ∈ Z2 ,
where v, a ∈ Z2 and k ∈ Z. In each of Lemma 3.1 and Lemma 3.5, three pairs
of vectors (vi , ai ) appear in such conditions, and these vectors have several suggestive properties. If we view each vector as a Gaussian integer Z[i] via the usual
identification x 7→ x1 + ix2 , then we have
v1 + v2 + v3 = 0
and a1 + a2 + a3 = 0.
In this section, we generalize these vectors, associating pairs (vi , ai ) of vectors to
each circle in B. Our calculations largely follow those of Stange [23], who, motivated
by data on our lattices LC for C ∈ B and Conway’s association of lattices to
quadratic forms [5], studied ways to associate lattices to circles in an Apollonian
circle packing.
4.2. Action of the Apollonian group. For the rest of this paper, we identify
C with R2 and Z[i] with Z2 in the usual way. Curvature coordinates are made
complex by writing C = (c, cz) ∈ R × C for a circle with radius c−1 and center
z ∈ C. Part of the band packing in complex curvature coordinates is shown in
Figure 4.1. As noted in the introduction, a pairwise-tangent triple (C1 , C2 , C3 ) is
related linearly to its Soddy circles C0 , C4 in curvature coordinates by
C0 + C4 = 2(C1 + C2 + C3 ).
This relation works also for lines, with the convention that the curvature coordinates
of a line ` are (0, z) where z is the unit normal vector to the line, oriented away
from the component of R2 \` containing the other circles in the triple. In particular,
all lines the circle packing B have coordinates (0, −1) or (0, 1), depending on which
side of the line the quadruple under consideration lies in.
A Descartes quadruple is a list of four circles such that any three form a pairwisetangent triple. As any pairwise tangent triple of circles has exactly two Soddy circles, any pairwise tangent triple of circles can likewise be completed to exactly
two Descartes quadruples, up to permutation. We call a Descartes quadruple
(C0 , C1 , C2 , C3 ) ∈ B 4 proper if the curvatures satisfy c0 > max{c1 , c2 , c3 } and points
of tangency between C0 and C1 , C2 , C3 are clockwise around C0 , or if
C0 = (1, 1 + 2z) and (Ci , Cj , Ck ) = ( (1, 1 + 2z + 2i), (0, 1), (0, −1) )
for z ∈ Z[i] and (i, j, k) a rotation of (1, 2, 3). We call the circle C0 the child and
the circle C1 , C2 , C3 the parents.
Note that, if (C0 , C1 , C2 , C3 ) ∈ B 4 is a proper Descartes quadruple, then so is the
parent rotation (C0 , C2 , C3 , C1 ) and the successor (2(C0 +C2 +C3 )−C1 , C0 , C2 , C3 ).
Each circle C0 ∈ B determines a Descartes quadruple (C0 , C1 , C2 , C3 ) up to parent
rotation. Moreover, any pairwise tangent triple (C1 , C2 , C3 ) ∈ B 3 can be completed
to a proper Descartes quadruple (C0 , C1 , C2 , C3 ) in a most one way.
INTEGER SUPERHARMONIC MATRICES
17
1, 1 + 2i
64,
1 + 112i
49,
1 + 84i
36,
1 + 60i
64,
127 + 112i
49,
97 + 84i
36,
71 + 60i
25,
1 + 40i
25,
49 + 40i
84,
7 + 132i
84,
161 + 132i
81,
1 + 126i
81,
161 + 126i
16,
1 + 24i
16,
31 + 24i
52,
7 + 76i
52,
97 + 76i
49,
1 + 70i
49,
97 + 70i
72,
17 + 96i
72,
127 + 96i
28,
7 + 36i
28,
49 + 36i
57,
17 + 72i
81,
17 + 102i
57,
97 + 72i
96,
31 + 120i
25,
1 + 30i
9,
17 + 12i
81,
145 + 102i
64,
127 + 80i
96,
161 + 120i
76,
7 + 92i
76,
145 + 92i
25,
49 + 30i
49,
1 + 56i
49,
97 + 56i
64,
31 + 72i
64,
97 + 72i
81,
1 + 90i
81,
161 + 90i
33,
17 + 36i
88,
49 + 96i
88,
127 + 96i
97,
49 + 102i
4, 1 + 4i
33,
49 + 36i
97,
145 + 102i
73,
49 + 76i
12,
7 + 12i
24,
17 + 24i
60,
49 + 60i
84,
71 + 84i
84,
97 + 84i
60,
71 + 60i
40,
49 + 40i
73,
49 + 70i
24,
31 + 24i
12,
17 + 12i
73,
97 + 70i
97,
49 + 92i
33,
17 + 30i
4, 7 + 4i
73,
97 + 76i
40,
31 + 40i
97,
145 + 92i
88,
49 + 80i
88,
127 + 80i
33,
49 + 30i
81,
1 + 72i
81,
161 + 72i
64,
31 + 56i
64,
97 + 56i
49,
1 + 42i
49,
97 + 42i
25,
1 + 20i
76,
7 + 60i
76,
145 + 60i
64,
1 + 48i
96,
31 + 72i
81,
17 + 60i
9,
1 + 6i
25,
49 + 20i
64,
127 + 48i
96,
161 + 72i
57,
17 + 42i
81,
145 + 60i
57,
97 + 42i
28,
7 + 20i
28,
49 + 20i
72,
17 + 48i
72,
127 + 48i
0, 1
0, −1
9,
1 + 12i
64,
1 + 80i
9,
17 + 6i
49,
1 + 28i
49,
97 + 28i
52,
7 + 28i
52,
97 + 28i
16,
1 + 8i
16,
31 + 8i
81,
1 + 36i
81,
161 + 36i
84,
7 + 36i
84,
161 + 36i
25,
1 + 10i
25,
49 + 10i
1, 1
36,
1 + 12i
49,
1 + 14i
64,
1 + 16i
36,
71 + 12i
49,
97 + 14i
64,
127 + 16i
Figure 4.1. The Apollonian Band packing B0 with complex curvature coordinates, in the region 0 < Im(z) < 2. The circles
tangent to the line (0, −1) are Ford circles.
4.3. Lattice vectors. We now assign vectors v(C, C 0 ), a(C, C 0 ) ∈ Z[i] to each
pair of tangent circles C, C 0 ∈ B. In analogy to the two degenerate cases, the
vectors v(C, C 0 ) and a(C, C 0 ) will generate periodicity conditions of the odometers
associated to C and C 0 .
To describe our recursive construction, let us call a Descartes quadruple semiproper if it is either proper or a parent rotation of
(C0 , C1 , C2 , C3 ) = ( (1, 1 + 2z), (1, 1 + 2z + 2i), (0, 1), (0, −1) )
for some z ∈ Z[i]. These latter quadruples are associated to the large circles in
B and fail to be proper because one of the parents has the same curvature as the
child. Note that, modulo parent rotations, the set of semi-proper quadruples is a
directed forest in which each node has exactly one parent and three children. We
induct along this tree structure to define our vectors.
For each semi-proper Descartes quadruple (C0 , C1 , C2 , C3 ) we define v(Cj , Ci )
and a(Cj , Ci ) for all rotations (i, j, k) of (1, 2, 3). For the base case, we consider the
clockwise Descartes quadruple
(C0 , C1 , C2 , C3 ) = ( (1, 1 + 2z), (1, 1 + 2z + 2i), (0, 1), (0, −1) )
for z ∈ Z[i],
and set
v(C3 , C2 ) = 0
a(C3 , C2 ) = 1
v(C2 , C1 ) = 1
a(C2 , C1 ) = z
v(C1 , C3 ) = −1
a(C1 , C3 ) = −1 − z.
For the induction step, we fix a semi-proper Descartes quadruple (C0 , C1 , C2 , C3 )
and suppose that v(Cj , Ci ) and a(Cj , Ci ) have been defined for all rotations (i, j, k)
18
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
of (1, 2, 3). For the successor quadruple (2(C0 +C2 +C3 )−C1 , C0 , C2 , C3 ), we define
v(C2 , C0 ) = v(C2 , C1 ) − iv(C3 , C2 )
v(C0 , C3 ) = v(C1 , C3 ) + iv(C3 , C2 )
a(C2 , C0 ) = a(C2 , C1 ) + ia(C3 , C2 )
a(C0 , C3 ) = a(C1 , C3 ) − ia(C3 , C2 ),
so that v(Cj , Ci ) and a(Cj , Ci ) are defined for all rotations (i, j, k) of (0, 2, 3). Since
every pair of tangent circles C, C 0 ∈ B can be completed to a proper Descartes
quadruple of the form (C0 , C, C 0 , C3 ), this recursive construction generates vectors
for all pairs of tangent circles in B. Since the quadruple of the big circle (1, 1 + 2z)
is a successor of the quadruple of (1, 1 + 2z + 2i), we must also check everything is
well-defined, but this is immediate from the definition.
Lemma 4.1. If (C0 , C1 , C2 , C3 ) ∈ B 4 is a proper Descartes quadruple and we write
Ci = (ci , ci zi ), vij = v(Ci , Cj ), and aij = a(Ci , Cj ), then the following hold.
v32 + v13 + v21 = 0
a32 + a13 + a21 = 0
(4.1a)
v10 = v13 − iv21
a10 = a13 + ia21
(4.1b)
v01 = iv10
a01 = −ia10
(4.1c)
2
v32
2v32 a32 = c3 z3 + c2 z2
(4.1d)
= c3 c2 (z3 − z2 )
v̄13 v21 + v13 v̄21 = −2c1
(4.1e)
The properties (4.1a), (4.1b), and (4.1c) give inductive relationships among the
vectors. The property (4.1d) expresses the vectors v(C, C 0 ) and a(C, C 0 ) up to sign
in terms of the circles C, C 0 ∈ B. Finally, (4.1e) implies that the determinant of of
the lattice generated by v31 and v21 is c1 .
Proof of Lemma 4.1. The inductive construction gives (4.1a)–(4.1c) immediately.
To check (4.1d), we first observe that it holds for the base case and its immediate
successors. By induction, it suffices to assume (4.1d) holds for the proper Descartes
quadruples (C0 , C1 , C2 , C3 ), (C1 , C4 , C2 , C3 ) ∈ B 4 and check the conditions for each
of the three successors of (C0 , C1 , C2 , C3 ). We write wi = ci zi and compute
2
v10
= (v13 − iv21 )2
= (v13 − v12 )2
2
2
= 2v13
+ 2v12
− (v13 + v12 )2
2
2
2
= 2v13
+ 2v12
− v14
= 2(c3 w1 − c1 w3 ) + 2(c2 w1 − c1 w2 ) − (c4 w1 − c1 w4 )
= (2(c1 + c2 + c3 ) − c4 )w1 − c1 (2(w1 + w2 + w3 ) − w4 )
= c0 w1 − c1 w0 ,
INTEGER SUPERHARMONIC MATRICES
19
using the induction hypotheses and the linear Soddy relation. We also compute
v10 a10 = (v13 − iv21 )(a13 + ia21 )
= v13 a13 + v21 a21 + i(v13 a21 − v21 a13 )
= v13 a13 + v21 a21 − v13 a12 − v12 a13
= 2v13 a13 + 2v21 a21 − (v13 + v12 )(a13 + a12 )
= 2v13 a13 + 2v21 a21 − v14 a14
= 21 (w1 + 2(w1 + w2 + w3 ) − w4 )
= 21 (w1 + w0 ).
2
The relations vi0
= ci c0 (zi − z0 ) and 2vi0 ai0 = ci zi + c0 z0 for i = 2, 3 follow by
analogous computations.
2
The relation (4.1e) can be obtained as follows. Observe that vij
= ci cj (zi − zj )
2
implies |vij | = ci + cj and therefore
2
2
v13 v21 (v̄13 v21 + v13 v̄21 ) = (c1 + c3 )v21
+ (c2 + c1 )v13
2
2
= c1 (v21
+ v13
) + c3 (c1 w2 − c2 w1 ) + c2 (c3 w1 − c1 w3 )
2
2
= c1 (v21
+ v13
) − c1 (c3 w2 − c2 w3 )
2
2
2
= c1 (v21
+ v13
− v32
)
= −2c1 v21 v13 .
Since at most one of the Ci is a line when we are not in the base case, we have
vij 6= 0 and thus obtain the next relation.
Given a circle C0 in the band packing B with the proper Descartes quadruple
(C0 , C1 , C2 , C3 ), we define the lattice
ΛC0 = Zv(C1 , C0 ) + Zv(C2 , C0 ) + Zv(C3 , C0 )
P3
(any two of the three summands suffice since i=1 v(Ci , C0 ) = 0). Next we compare
ΛC with the lattice LC of (1.4).
Lemma 4.2. ΛC ⊆ LC for all C ∈ B.
Proof. If C0 ∈ B has radius 1, then ΛC0 = LC0 = Z2 and there is nothing to
prove. Otherwise, we may assume that C0 is part of a proper Descartes quadruple
(C0 , C1 , C2 , C3 ) ∈ B 4 . Using Lemma 4.1, we compute for i = 1, 2, 3
−1
ai0 = 12 (ci zi + c0 z0 )vi0
−1
2
= 12 (c−1
0 vi0 + (ci + c0 )z0 )vi0
1
= 21 c−1
0 vi0 + 2 z0 vi0
(4.2)
= AC0 vi0 .
Since AC0 vi0 = ai0 ∈ Z[i] we conclude that vi0 ∈ LC0 for i = 1, 2, 3.
In fact ΛC = LC for all C ∈ B, but our proof of the inclusion LC ⊆ ΛC uses our
construction of the odometers gC , and is thus postponed to the end of Section 9.
Observe that (4.1b) can be rewritten as v0i = vji ± vki , where (i, j, k) is a
rotation of (1, 2, 3) and sign of vki depends on whether or not Ci is a parent of Ck .
Inductively, this implies that a vector v(C, C0 ) lives in the lattice of the circle C0 :
20
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
Lemma 4.3. Given a proper Descartes quadruple (C0 , C1 , C2 , C3 ) ∈ B 4 and any
C ∈ B which is tangent to and smaller than C0 , we have that v(C, C0 ) ∈ ΛC0 . Finally, we check that our general lattice vector construction agrees with the
degenerate cases.
Proposition 4.4. Every Ford circle C0 = (q 2 , 1+2pqi) is part of a proper quadruple
(C0 , C1 , C2 , C3 ), where C1 = (q12 , 1 + 2p1 q1 i), C2 = (q22 , 1 + 2p2 q2 i), and C3 =
(0, −1). Moreover, we have
v(C1 , C0 ) = q + q1 i
a(C1 , C0 ) = p1 + pi
v(C2 , C0 ) = −q + q2 i
a(C2 , C0 ) = p2 − pi
v(C3 , C0 ) = −qi
a(C3 , C0 ) = −pi,
which are exactly the vectors appearing in Lemma 3.1.
Proof. That (C0 , C1 , C2 , C3 ) is proper follows from the discussion in Section 3.2.
Since (4.1a) and (4.1d) determines the tuple (v10 , v20 , v30 , a10 , a20 , a30 ) up to sign,
it is enough to show v30 = −qi. Using (4.1b) and (4.1c), we inductively compute
v30 = v32 − iv13 = v32 + iv13 = −iq2 − iq1 = −iq.
We conclude after reading off the base case C0 = (1, 1 + 2) from the beginning of
this section.
Proposition 4.5. Every diamond circle C0 = (2k(k + 1), 2k 2 − 1 + 2k(k + 1)i) is
part of a proper quadruple (C0 , C1 , C2 , C3 ), where C1 = (1, 1 + 2i), C2 = (1, 1), and
C3 = (2(k − 1)k, 2(k − 1)2 − 1 + 2(k − 1)ki). Moreover, we have
v(C1 , C0 ) = k + 1 + ki
a(C1 , C0 ) = k + i
v(C2 , C0 ) = −k − 1 + ki
a(C2 , C0 ) = −ki
v(C3 , C0 ) = −2ki
a(C3 , C0 ) = −k + (k − 1)i,
which are exactly the vectors appearing in Lemma 3.5.
Proof. That (C0 , C1 , C2 , C3 ) is proper follows from the discussion in Section 3.3.
As in the previous proposition, it is enough to show v(C1 , C0 ) = k + 1 + ki. Using
(4.1b) and (4.1c), we inductively compute
v10 = v13 − iv21 = (k + (k − 1)i) + (1 + i) = k + 1 + ki.
We conclude after reading off the base case C0 = (4, 1 + 4i) from the previous
proposition.
4.4. Symmetry reduction. The band packing B is invariant under the operations
(c, z) 7→ (c, −z)
(c, z) 7→ (c, z̄)
(c, z) 7→ (c, z + 2ci)
(c, z) 7→ (c, z + 2c).
These operations can be extended to the vectors v(C, C 0 ) and a(C, C 0 ) in the obvious
way. For example, if we apply the shift (c, z) 7→ (c, z + 2cw) for some w ∈ Z[i], then
v(C, C 0 ) is unchanged while a(C, C 0 ) is replaced by a(C, C 0 ) + cc0 v(C, C 0 )−1 .
We can extend the results of Section 3 to the orbit of the Ford and diamond
circles under these symmetries. For example, suppose that C = (c, z) is a Ford
INTEGER SUPERHARMONIC MATRICES
21
circle and we want to construct and odometer for the shifted circle C 0 = (c, z +2cw)
for w = a + bi ∈ Z[i]. Observe that
a b
0
AC − AC =
b −a
and that the function hw : Z2 → Z given by
a
a
hw (x) = x1 (x1 + 1) − x2 (x2 + 1) + bx1 x2
2
2
satisfies ∆hw ≡ 0. In particular, gC 0 = gC + hw is an odometer for C 0 .
5. A topological lemma
In the course of our general tile construction, it is necessary to translate local knowledge of tile compatibility to global knowledge regarding the intersection
structure of the collection of tiles. This section concerns a technical lemma which
is our primary tool for checking that certain collections of tiles form tilings using
local conditions.
We continue to identify Z[i] with Z2 in the natural way. Thus a tile T is a set of
squares sx = {x, x + 1, x + i, x + 1 + i} ⊆ Z[i], where I(T ) (adapted from (1.6) to C
in the obvious way) is a topological disc. We let Z[i] inherit the standard degree4 square lattice graph of Z2 . For any tile, define F (T ) to be the subgraph of Z[i]
induced by the union of squares sx ∈ T . For two tiles T1 , T2 write F (T1 ) ∩ F (T2 ) for
the induced subgraph on the intersection of the vertex sets of F (T1 ) and F (T2 ). In
an abuse of terminology, we say that tiles T1 and T2 intersect if F (T1 ) ∩ F (T2 ) 6= ∅,
and overlap if T1 ∩ T2 6= ∅. If T1 ∩ T2 = ∅, they are non-overlapping.
Lemma 5.1. Suppose T is an (infinite) collection of tiles and E is a set of twoelement subsets of T satisfying the following four hypotheses.
(1) The graph G = (T , E) is a 3-connected planar triangulation.
(2) T and
P E are invariant under translation by some full-rank lattice L ⊆ Z[i],
and T ∈T /L |T | = | det L|.
(3) If {T1 , T2 } ∈ E then the intersection F (T1 ) ∩ F (T2 ) contains at least 2
vertices.
(4) We can select for each face F = {T1 , T2 , T3 } of G a point ρ(F ) ∈ F (T1 ) ∩
F (T2 ) ∩ F (T3 ) such that, for the face F 0 = {T1 , T2 , T4 } of G, ρ(F ) and
ρ(F 0 ) lie on some path contained in F (T1 ) ∩ F (T2 ).
Then T is a tiling. Moreover, if T1 , T2 ∈ T and T1 and T2 intersect, then {T1 , T2 } ∈
E, and F (T1 ) ∩ F (T2 ) is a path joining ρ(F ) and ρ(F 0 ) for the faces F, F 0 with
F ∩ F 0 = {T1 , T2 }.
The proof is essentially a homotopy argument. Before commencing, recall that
the winding number of a closed walk in Z[i] about a point z ∗ ∈ Z[i]∗ is the number
of times the walk circles the point z ∗ counterclockwise; thus in terms of the vertices
z1 , z2 , . . . , zt = z1 of the walk, the winding number is given by
t−1
zi+1 − z ∗
1 X
∗
arg
.
wind(z1 , . . . , zt ; z ) :=
2π i=1
zi − z ∗
We say that the closed walk W encloses a point z ∗ if the winding number of W
about Z ∗ is nonzero; similarly, we say that W encloses a square sx = {x, x + 1, x +
i, x + 1 + i} ⊆ Z[i] if the winding number of the walk about s∗x = x + 21 + 2i is
22
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
nonzero. Note that the definition of a tile (via (1.6)) implies that any tile T is a
set of squares enclosed by a simple cycle ∂T in Z2 , which is the set of points and
edges of T which each also lie in a square sx ∈
/ T.
Proof of Lemma 5.1. Our first goal is to show that T is a tiling. Note that it
suffices to show that every square sx = {x, x + 1, x + i, x + 1 + i} lies in some tile
in T , since hypothesis (2) implies that the average number of tiles a square lies in
is 1. The idea is to use the periodicity to draw a large cycle that surrounds a given
square sx , and then use the graph structure to contract this cycle to the boundary
of a single tile.
We fix an L-periodic drawing of G = (T , E), and work with the dual graph G∗ ,
whose vertex set F is the set of triangles in the graph (T , E). For F ∈ F, we have
that ρ(F ) is a point in the intersection of the three tiles in F , and for adjacent
F, F 0 , let ρ(F, F 0 ) be a choice of path from ρ(F ) to ρ(F 0 ) in the intersection of
the two tiles in F ∩ F 0 , guaranteed to exist by hypothesis (4). Given a sequence
F 0 , F 1 , . . . , F n = F 0 with |F i ∩ F i+1 | = 2 (indices evaluated modulo n), we define
the closed walk `(F 0 , . . . , F n ) in Z2 as the consecutive concatenation of the paths
ρ(F i , F i+1 ) (0 ≤ i < n).
Since the graph (T , E) is connected and periodic under a nontrivial lattice, we
can, given any finite set Z ⊆ Z[i]∗ , find a cycle F 0 , . . . , F n = F 0 in (T , E) which
wraps around each point z ∗ ∈ Z, just in the sense that wind(ρ(F 1 ), . . . , ρ(F n ); z ∗ ) 6=
0 for all z ∗ ∈ Z. Since hypothesis (2) also implies that the tiles T ∈ T are of
bounded size, this means that given the point s∗x ∈ Z[i]∗ , we can, in fact, find a
cycle C = F 0 , F 1 , . . . , F n = F 0 such that `(C) encloses s∗x and so sx . Letting K
denote the set of faces of G∗ in the region of the plane bounded by C in our fixed
embedding (so, each element of K corresponds to some T ∈ T ), we choose this C
such that |K| is as small as possible, subject to the condition that `(C) encloses s∗x .
Note that if |K| = 1, then sx is indeed covered by T , since then all vertices in
`(C) belong to a single tile, which is simply connected and would thus cover sx .
Moreover, any element of K whose intersection with C is disconnected is a cutvertex of K, when K is viewed as a subgraph of the original graph G. Since not
every vertex in a finite graph can be a cut-vertex, we may assume without loss of
generality that there is a cycle
C 00 = F n , F n−1 , . . . , F ` , E1 , . . . , Et
for ` > 0 which is a member of K, and where no Ei lies on C. We now consider the
closed walk `(C 0 ) in Z[i], where C 0 is the cycle
C 0 = F 0 , F 1 , . . . , F ` , E1 , E2 , . . . , Et , F n
in G∗ . Note that the region bounded by C 0 has exactly one fewer face than that
bounded by C. We will show that if `(C) enclosed sx , then so must `(C 0 ), contradicting minimality of C.
To begin, note that the face C 00 of the dual graph G∗ corresponds to some vertex
T ∈ T , which completely contains the walk `(C 00 ). Thus, `(C 00 ) does not enclose
sx unless sx lies in T . Finally, the winding number about any square sx of the loop
`(C 0 ) is the same as the winding number about the square sx of the concatenation
of `(C) with `(C 00 ). In particular, since sx is not enclosed by `(C 0 ), the winding
number of `(C) and `(C 00 ) about sx must be equal. Thus `(C 00 ) also encloses sx ,
contradicting the minimality of C with this property.
INTEGER SUPERHARMONIC MATRICES
23
Having shown that T is a tiling, we next wish to show that if F (T1 )∩F (T2 ) 6= ∅,
then F (T1 ) ∩ F (T2 ) is a path in Z[i] from ρ(F ) to ρ(F 0 ) for the faces F, F 0 whose
intersection is the pair {T1 , T2 }. We begin by showing that the intersection is a
path.
If it is not, there are paths P1 ⊆ ∂T1 and P2 ⊆ ∂T2 with the same pair of
∗
endpoints, whose concatenation C is a cycle enclosing a region S ∗ of Z[i] disjoint
∗
∗
from T1 and T2 . Among all possible pairs T1 , T2 , we may assume we have chosen
such that |S ∗ | is as large as possible (note that there is some absolute bound on
|S ∗ |, since, for example, hypothesis (2) implies that tiles have bounded size).
We let TS denote the tiles in the region bounded by C. Since (T , E) is 3connected, there must be at least 3 tiles in T \ TS which are adjacent to tiles in TS .
By hypothesis (3), such a tile T must have the property that it shares two vertices
with some tile in TS ; however, the only candidate points to be shared between a
tile T ∈
/ TS , T 6= T1 , T2 , and a tile in TS are the 2 common endpoints of P1 and
P2 , and for C to be a (simple) cycle there is, for each of these two endpoints, at
most one square of Z[i] containing the point and lying outside C and outside of the
tiles T1 , T2 . In particular, there must be exactly three tiles in T \ TS adjacent to
tiles in TS ; namely, T1 , T2 , and a third tile T3 which includes both endpoints of the
paths P1 , P2 . But now either the pair {T3 , T2 } or the pair {T3 , T1 } contradict the
maximality of the choice of S ∗ . Thus F (T1 ) ∩ F (T2 ) is indeed a path.
ρ(F ) and ρ(F 0 ) both lie in F (T1 ) ∩ F (T2 ). If they are not the endpoints of
the path, then, without loss of generality, let e1 , e2 be the two edges of the path
F (T1 ) ∩ F (T2 ) which are incident with ρ(F ). Since T1 and T2 are nonoverlapping,
for each i = 1, 2, ei lies in one square from T1 and one square of T2 . Moreover, their
shared endpoint ρ(F ) lies also in a third tile T3 (which again, is nonoverlapping
with T1 , T2 ). In particular, either T1 or T2 must contain two diagonally opposite
squares about ρ(F ) without containing the other two squares, contradicting the
definition of a tile.
Finally, we wish to show that if F (T1 )∩F (T2 ) 6= ∅, then {T1 , T2 } ∈ E. We do this
by giving a suitable plane drawing of the graph G = (T , E). For each tile T ∈ T ,
we draw a vertex vT corresponding to T at some point of the dual T ∗ . Identifying
C with the Euclidean plane, let s̄x = {x + s + ti | 0 ≤ s, t ≤ 1} ⊆ C. Since we know
that tiles T, T 0 ∈ T which are adjacent in G must share an edge of Z[i], we can
draw a curve from vT to vT 0 such that every point in the curve lies in the interior
of s̄x ∪ s̄y where sx , sy each lie in T or T 0 ; in particular, we can draw all edges of G
such that they are pairwise nonintersecting (except at shared endpoints) and such
that the edge from vT to vT 0 is disjoint from any s̄x for an sx not contained in T or
T 0 . With this drawing the curve CT in C corresponding to the cycle through the
neighbors of a tile T is disjoint from T , and bounds a region containing T . Since
CT and CT 0 bound disjoint regions of C when T, T 0 are nonadjacent in G, we see
that any nonadjacent tiles are nonintersecting.
6. Tiles
In this section, we associate to each circle C ∈ B a tile (unique up to translation)
which will be 90◦ symmetric and tile the plane with the lattice ΛC . Before beginning
our construction, we need a few additional definitions regarding tiles. We let c(T )
and |T | denote the centroid and area of a tile T , which are, respectively, the centroid
24
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
and area of the real subset I(T ) from (1.6). We say that T1 and T2 touch if they
are non-overlapping and ∂T1 ∩ ∂T2 is a simple path of Z2 with at least two vertices.
We say that three tiles form a touching triple of tiles if they are pairwise touching,
and share exactly one common boundary vertex.
It will be convenient to allow a degenerate case of our tile definition. Note that
if T = ∅, then the centroid c(T ) would be undefined. We will allow tiles T = ∅
to have F (T ) assigned as an arbitrary singleton from Z[i]; this singleton then gives
the centroid of T . If T is degenerate, we say that T, T 0 touch if F (T ) ⊆ ∂T 0 . We
emphasize that T \ T 0 = T whenever T 0 is degenerate.
Let a prototile T be a set of squares sx ⊆ Z[i], and, if empty, have F (T ) assigned
as a singleton in Z[i] (compared with the definition of a tile, we are dropping the
requirement that I(T ) is a topological disk ). We begin by recursively associating
a prototile to each C ∈ B; in Lemma 6.3 we will verify that these prototiles are in
fact tiles.
Definition 6.1. If (C0 , C1 , C2 , C3 ) ∈ B 4 is a proper Descartes quadruple, then a
set T0 of squares sx ⊆ Z[i] is a prototile for C0 if T0 has the tile decomposition
T0 = T1+ ∪ T1− ∪ T2+ ∪ T2− ∪ T3+ ∪ T3− ,
(6.1)
with F (Ti± ) ⊆ F (T0 ) even if Ti is degenerate, where, for each rotation (i, j, k) of
(1, 2, 3), Ti± is a prototile of Ci satisfying
c(Ti± ) − c(T0 ) = ± 21 (vkj − ivkj ).
(6.2)
where vij := v(Ci , Cj ).
The base cases are those circles in B which are not the first circle of any proper
Descartes quadruple: T0 is a prototile for C0 = (0, ±1) if T0 = ∅ and F (T0 ) = {x}
for any x ∈ Z[i], while T0 is a prototile for C0 = (1, 1 + 2z) if T0 = {sx }, for any
x, z ∈ Z[i].
Note that by induction, any circle in B has at most one prototile up to translation,
and any circle’s prototiles must necessarily be 180◦ symmetric.
When a prototile T for a circle C satisfies the definition of a tile (i.e., I(T ) is a
topological disk ), we say that T is a tile for C. Given a tile T0 for a circle C0 ∈ B
with c0 > 1, we say T is a subtile of T0 if T is one of the tiles in the decomposition
(6.1) for T0 . In the proof of Lemma 6.3, below, we will see that the decomposition of
T into subtiles is nonoverlapping, except for prescribed overlap between the largest
pair of subtiles.
Our construction in Section 3.2 recursively assigns tiles to each Ford circle with
decompositions
T0 = T1+ ∪ T1− ∪ T2+ ∪ T2−
where the Ti± ’s were constructions for the two Ford parents of C0 . As a general
circle in the Apollonian packing, the third parent of a Ford circle is a line (0, −1)
with the degenerate tile ∅; thus to see that T0 can be realized as a tile for C0
via Definition 6.1, it is only necessary to check, via Proposition 4.4, that assigning
F (T3± ) = c(T0 ) ± 21 (v12 − iv12 ) (from (6.2)) gives that F (T3± ) ⊆ F (T0 ).
The presence of the coefficient 21 in (6.2) means that even the existence of prototiles for general circles is not quite immediate.
Lemma 6.2. There is a prototile T0 for every C0 ∈ B.
INTEGER SUPERHARMONIC MATRICES
T3−
T1+
+
S2
Q−
3 Q+
4
T3−
T2+
−
S3
25
−
S3
+
S4
T2−
−
S2
−
+
S4
S3
Q+
2
T2+
+
S2
T3+
+
S4
Q+
3
T2−
−
S2
−
+
S4
S3
Figure 6.1. Constructing a
prototile from the decomposition of its largest parent tile.
Q−
2
T3+
Figure 6.2. The double decomposition of a tile.
Proof. If C0 is not equivalent to a Ford circle under the symmetries discussed in
Section 4.4, then it is a member of a proper Descartes quadruple (C0 , C1 , C2 , C3 ) for
Ci 6= (0, ±1) for each i, and (C1 , C4 , C2 , C3 ) is also a proper Descartes quadruple,
where C4 = 2(C1 + C2 + C3 ) − C0 is the Soddy precursor of C0 . By induction, C1
has a prototile T1 , which is either a translate of T1 = {s0 }, in which case a direct
application of Definition 6.1 verifies that T0 = {s−1−i , s−1 , s−i , s0 } is a prototile for
C0 , or otherwise is given as
T1 = S4+ ∪ S4− ∪ S2+ ∪ S2− ∪ S3+ ∪ S3− ,
where each Si± is a prototile for Ci such that
c(Si± ) − c(T1 ) = ± 12 (vkj − ivkj ),
for all rotations (i, j, k) of (4, 2, 3). In this latter case, we set
T1+ = T1 + v32 − iv32
T2+ = S2+ + v32
T3+ = S3+ − iv32
T1− = T1
T2− = S2− − iv32
T3− = S3− + v32 .
See Figure 6.1 for an illustration. The key point is that v32 ∈ Z[i], and thus that
each Ti± is a prototile for Ci , allowing us to define
T0 := T1+ ∪ T1− ∪ T2+ ∪ T2− ∪ T3+ ∪ T3− .
By the lattice rules, we compute that
c(Ti± ) − p0 = ± 21 (vkj − ivkj ),
if (i, j, k) is a rotation of (1, 2, 3), and p0 := c(T1− ) + 12 (v32 − iv32 ). Since each
Ti± is 180◦ symmetric, we must have p0 = c(T0 ). Thus T0 is a prototile for C0
according to Definition 6.1. Note that the condition that F (Ti± ) ⊆ T0 cannot fail,
since Ci 6= (0, ±1) for any i implies that no Ti± is degenerate.
Lemma 6.3, below, is the heart of our inductive construction, showing that prototiles for circles in B are in fact 90◦ symmetric tiles. Lemma 6.3 regards a proper
Descartes quadruple (C0 , C1 , C2 , C3 ), writing Ci = (ci , zi ) and vij = v(Ci , Cj ),
and where Ti denotes some prototile for Ci . For simplicity of notation, we assume
c1 ≥ c2 ≥ c3 , using the symmetries discussed in Section 4.4. A major goal of
the lemma is to understand the relationship between T0 and the tiles T0 ± v0i , for
i = 1, 2, 3. We do this by decomposing tiles, to deduce that tiles form a touching
26
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
3
R1
0
R3
2
R2
1
R2
T3−
1
R3
T1+
T2+
2
R1
0
R1
T2−
T1−
3
R3
T3+
3
R2
0
R2
2
R3
1
R1
Figure 6.3. Verifying a tiling by decomposing into parent tiles.
The large tile in the center is T0 .
triple as a consequence of the fact that some smaller tiles form a touching triple.
In particular, we write T0 as a union of Ti± as in (6.1) and then define
R10 = T1+ + v03
R20 = T2+ + v01
R30 = T3+ + v02
R11 = T1+ − v02
R21 = T2+ − v03
R31 = T3+ − v01
R12 = T1− − v03
R22 = T2− − v01
R32 = T3− − v02
R13 = T1− + v02
R23 = T2− + v03
R33 = T3− + v01 .
(6.3)
(See Figure 6.3.) When c1 > 1, Definition 6.1 implies that the subtiles T1± have
decompositions as in (6.1), and we write
−
+
−
+
−
T1+ = Q+
2 ∪ Q2 ∪ Q3 ∪ Q3 ∪ Q4 ∪ Q4 ,
T1− = S2+ ∪ S2− ∪ S3+ ∪ S3− ∪ S4+ ∪ S4− ,
±
where Q±
i and Si are each tiles for Ci , and (C1 , C4 , C2 , C3 ) is the proper Descartes
quadruple given by letting C4 = 2(C1 + C2 + C3 ) − C0 ∈ B be the Soddy precursor
of C0 . The double decomposition of T0 is then the collection of tiles
{T2+ , T2− , T3+ , T3− } ∪ {Si± | i = 2, 3, 4} ∪ {Q±
i | i = 2, 3, 4}
shown in Figure 6.2. Note that in the course of proving Lemma 6.3, we will show
+
that Q−
4 = S4 .
INTEGER SUPERHARMONIC MATRICES
27
Lemma 6.3. Let (C0 , C1 , C2 , C3 ) be a Descartes quadruple, let (i, j, k) indicate any
rotation of (1, 2, 3), and let vij = v(Ci , Cj ). Denoting the tiles of the decomposition
and double decomposition of T0 as above, the following properties hold:
(T1) T0 is a tile, and |T0 | = c0 . Moreover, T0 \ Ti± is a tile, which touches Ti± ,
for each i = 1, 2, 3.
(T2) T0 is 90◦ symmetric.
(T3) T0 , T0 + vi0 , and T0 − vj0 form a touching triple, provided c0 > 1.
(T4) T0 , T0 + vi0 , and Ti form a touching triple for c(Ti ) − c(T0 ) = 12 (vi0 + v0i ),
provided c0 > 1.
(T5) T0 , Ti , and Tj form a touching triple for c(Ti ) − c(T0 ) = 12 (vi0 + v0i ), and
c(Tj ) − c(T0 ) = 21 (vj0 − v0j ).
(T6) If c1 ≥ c2 > 1, then among other labeled tiles from Figure 6.3, the subtile
Ti± intersects only those which are drawn adjacent to it or with overlap.
+
+
−
Moreover, F (T1+ ) ∩ F (T2+ ) ⊆ F (Q+
2 ) ∩ F (T2 ), and F (T1 ) ∩ F (T1 ) ⊆
+
−
−
+
−
(F (Q2 ) ∩ F (S3 )) ∪ F (Q4 ) ∪ (F (Q3 ) ∩ F (S2 )).
Note that (T6) could be removed from Lemma 6.3 without compromising the
induction, but this technical information will be necessary for our use of tiles in the
construction of integer superharmonic representatives.
Before commencing with our inductive proof of Lemma 6.3 in the general case,
we use it to prove the following version of the tiling theorem from the introduction.
This will gives a very simple example of an application of Lemma 5.1.
Theorem 6.4. For every circle C ∈ B, there is a tile TC ⊆ Z2 with 90◦ rotational
symmetry, such that TC + ΛC is a tiling. Moreover, except when C has radius 1,
each tile in TC + LC borders exactly 6 other tiles.
Note that Theorem 1.3 follows from this and our confirmation in Theorem 9.5
that ΛC = LC .
Proof of Theorem 1.3. In the case where C ∈ B has curvature 1, the lattice ΛC
is 2 · Z2 and the corresponding tile T is simply a single square sx . Thus we let
T = T0 be the tile for a circle C0 in a Descartes quadruple (C0 , C1 , C2 , C3 ) with
c0 > 1, and write vij for the vectors v(Ci , Cj ). We consider a Z2 -periodic planar
graph G whose vertices correspond to the tiles T0 + ΛC0 in the tiling, where two
vertices are adjacent if the corresponding tiles differ by a vector v0i . (T3) and the
area condition from (T1) now allows us to verify all the hypotheses of Lemma 5.1.
Lemma 5.1 implies that each square of Z2 lies in exactly one tile among T0 + ΛC0
(indeed, T0 + ΛC0 is a tiling) and that all tile intersections correspond to edges in
our graph, which has degree 6.
The rest of this section is devoted to the proof of Lemma 6.3. Since we have
already addressed the case of Ford circles and diamond circles, we may assume that
(C0 , C1 , C2 , C3 ) ∈ B is a proper Descartes quadruple, such that no Ci is a line, and
at most one Ci has curvature ci = 1. In particular, c0 > 4, and, rotating the parents
(C1 , C2 , C3 ) and possibly conjugating the the original tuple using the symmetries
in Section 4.4, we may assume C3 is a parent of C2 and C2 is a parent of C1 so that
c1 > c2 ≥ 4. In particular, we also have that (C1 , C4 , C2 , C3 ) and (C2 , C3 , C5 , C6 )
are both proper Descartes quadruples for some C5 , C6 ∈ B, and by induction, we
may assume Lemma 6.3 holds for both of these quadruples.
28
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
For the sake of clarity and brevity, Claims 6.5 and 6.6 below are stated with the
aid of Figure 6.3 and Figure 6.2, respectively. For the purposes of the statements of
±
the claims, tiles S1 , S2 , S3 (from among the Ti± and Rij , or Ti± , Q±
i and Si labeled
in the figure) are considered to be drawn adjacently if their corresponding regions
in the figure overlap or share some portion of their boundaries as drawn. (T1− and
T1+ in Figure 6.3 are the only tiles drawn with overlap.) In the course of our proof
of Lemma 6.3, we will be verifying that the basic tile layout in the figure is correct.
Claim 6.5. If (S1 , S2 , S3 ) is a triple of adjacent labelled tiles in Figure 6.3, not
all of which are contained in T0 , then the Si form a touching triple, unless |S1 | =
|S2 | = |S3 | = 1, in which case the Si ’s are pairwise intersecting squares.
Proof of claim. To prove this claim, we use the lattice rules (4.1) to simplify the
differences between centers of adjacent tiles. One can check that adjacent tiles for
Ci are related by ±vji for some j < i. One can also check that, when i 6= j,
adjacent tiles for Ci and Cj are related by 12 is (vij + vji ) for s = 0, 1, 2, 3. From
this, we see that all of the triples (S1 , S2 , S3 ) from the claim fall into one of the
inductive versions of (T4) or (T5) for (C1 , C4 , C2 , C3 ) or (C2 , C3 , C5 , C6 ) so long as
|Si | 6= 1 for some i. The case where each Si is a tile for a circle of curvature 1 is
easily checked by hand.
To analyze touching tiles within T0 , we need to make use of the double decomposition. Figure 6.2’s depiction of the double decomposition is a bit deceptive,
however, as it shows S4± smaller than S3± and S2± , even though the size of S4± relative to S2± , S3± is unknown at this point. For example, the tiles S2− and S3− may
or may not touch, depending on the relative size S4+ .
The following claim is sufficient for our purposes, however, and does not depend
on the relative size of C4 to C2 and C3 :
Claim 6.6. S4+ = Q−
4 , and if (R1 , R2 , R3 ) is a triple of adjacent labelled tiles
in Figure 6.2, not all of which are contained in a single T1± , then the Ri form a
touching triple of tiles.
Proof of claim. Again we use the lattice rules (4.1) to simplify the differences between centers of adjacent tiles. In particular, we verify that S4+ = Q−
4 , and
that each triple covered by the claim is a case of (T4) or (T5) for the quadruple (C2 , C3 , C5 , C6 ).
The information from the double decomposition is not enough for us to claim
yet that, e.g., T1+ and T2+ touch (in particular, that they intersect only on their
boundaries), as we have not analyzed the topological relationships among all the
tiles in the double decomposition from Figure 6.2, which would be quite cumbersome
in light of the unknown relative size of the circle C4 . To proceed further at this
point will require Lemma 5.1.
In principle, we would like to apply Lemma 5.1 to the set TL of translates of the
tiles Ti± by the lattice L generated by {v01 , v02 , v03 }. The overlap of tiles T1+ and
T1− disallows this, however, as the second part of hypothesis 2 of the Lemma will
not be satisfied.
To work around this, we define T̃L by replacing each translate of T1+ by the
corresponding translate of T1+ \ Q−
4 , which, by induction, is a tile by (T1). We
consider the graph G whose vertex set is in correspondence with TL , and, also,
INTEGER SUPERHARMONIC MATRICES
29
therefore, T̃L . For any vertex v ∈ V (G), we write T (v) for the corresponding tile
in TL , and T̃ (v) for the corresponding tile in T̃L . (Note that unless T (v) is a Ltranslate of T1+ , we have that T (v) = T̃ (v).) A pair of vertices {u, v} is joined by
an edge in G if T (u) and T (v) are translates by a common vector in L of a tile pair
drawn adjacently or with overlap in Figure 6.3. Viewed as an abstract graph, G is
easily seen to have a drawing as a Z2 -periodic planar triangulation.
We let E and Ẽ be the sets of pairs {T (u), T (v)} and {T̃ (u), T̃ (v)}, respectively,
for {u, v} ∈ E(G). We claim that the graph (T̃L , Ẽ) satisfies the hypothesis of
Lemma 5.1. Hypothesis 1 is immediate, as is the first part of hypothesis 2. For the
second part of hypothesis 2, observe that (using (1.7)) we have
X
|T | = 2c1 + 2c2 + 2c3 − c4 = c0 = | det ΛC0 |
T ∈T̃L /L
It remains to verify hypotheses 3 and 4. We will need the following:
−
+
Claim 6.7. If sx ∈
/ T1− ∪ T1+ , then sx ∩ F (Q−
4 ) = ∅, unless |T3 | = |T2 | = 1.
−
+
Proof. By (T1), T1+ \ Q−
4 and T1 \ S4 are both tiles, which each touch T4 :=
−
+
Q4 = S4 . Thus neither of these tiles shares any squares with S4+ , and unless
|T3− | = |T2+ | = 1, Claim 6.6 or inductively, by Lemma 6.3, give that the intersection
−
+
2
2
of T1+ \ Q−
4 and T1 \ S4 contains edges of Z . Thus, the union S ⊆ Z of these
∗
tiles satisfies that S is connected; since S is centrally symmetric about the center
of S4+ , we have as a consequence that no square of S4+ can intersect any square
outside of S.
Consider now hypothesis (3) of Lemma 5.1. For any pair T̃ (u), T̃ (v) in Ẽ, we
have from the definition of G that the pair T (u), T (v) are drawn adjacently or
with overlap in Figure 6.3. Claims 6.5 and 6.6 now imply that F (T (u)) ∩ F (T (v))
contains at least two vertices. Finally, Claim 6.7 implies that F (T̃ (u)) ∩ F (T̃ (v)) =
F (T (u)) ∩ F (T (v)), unless, up to symmetry, we have T (u) = T1+ , T (v) = T1− . In
this case, however, Claim 6.6 implies directly that F (T̃ (u)) ∩ F (T̃ (v)) contains at
least 2 vertices.
To check hypothesis 4 for the graph (T̃ , Ẽ), we warm up by examining this for
the graph (T , E), where it also holds. We choose an assignment of the points ρ(F )
for faces F of G. For any face F = {u, v, w} whose three corresponding tiles form
a triple covered by Claim 6.5, we must choose ρ(F ) ∈ Z[i] to be the unique point
in the three-way intersection of the footprints of the tiles. Remaining faces F are
those whose corresponding tile triple lies entirely within T0 : i.e., the triple is either
(T1± , T2± , T3∓ ) (T1± , T1∓ , T2± ). In the first case, we use Claim 6.6 to choose (without
loss of generality) ρ({T1+ , T2+ , T3− }) to be the unique point in the intersection of the
+
−
touching triple Q+
2 , T2 , T3 . In the second case, we use Claim 6.6 to choose (without
+
loss of generality) ρ({T1 , T1− , T2+ }) to be the unique point in the intersection of the
+
−
touching triple Q+
2 , T2 , S3 .
With this choice for the ρ(F )’s, we see that for any adjacent F, F 0 corresponding
to tiles from Figure 6.3 other than the pair of faces corresponding to the tile triples
{T1+ , T1− , T2+ },
0
{T1+ , T1− , T2− }
(6.4)
that the points ρ(F ), ρ(F ) are joined by a simple path given as the intersection of
a single pair of tiles F (U 1 ) ∩ F (U 2 ) known to touch either by Claim 6.5 or Claim
6.6. This would verify hypothesis 4 of Lemma 5.1 in the graph (T , E) for these pairs
30
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
of faces. Note now that Claim 6.7 implies that the points ρ(F ) selected above also
lie in the three way intersections F (T̃ (u)) ∩ F (T̃ (v)) ∩ F (T̃ (w)) for {u, v, w} = F ,
and, moreover, that the simple paths in the intersections of the footprints of pairs
of tiles T (u), T (v) used above remain in the intersection F (T̃ (u)) ∩ F (T̃ (v)). In
particular, we have verified that with this assignment of ρ(F )’s, the condition hypothesis 4 holds also in the graph (T̃ , Ẽ) for all faces other than those corresponding
to translates of triples
−
+
{T1+ \ Q−
4 , T1 , T2 },
−
−
{T1+ \ Q−
4 , T1 , T2 },
and we deal with this final case separately, now.
+
Let L+ be the set of edges of ∂Q−
4 whose two incident squares both lie in T1 , and
−
−
−
define L similarly. Note that L is the central reflection through c(Q4 ) of L+ , and
+
that Claim 6.7 implies that L+ ∪ L− equals the edge-set of ∂Q−
4 . We claim that L
+
is the edge-set of a path, which suffices for us, since then, letting V (L ) denote the
set of vertices incident with edges in L+ , we have that F (S3− ) ∩ F (S4+ ) ∩ F (Q+
2)⊆
+
−
V (L+ ) ∩ V (L− ) and F (S2− ) ∩ F (S4+ ) ∩ F (Q+
)
⊆
V
(L
)
∩
V
(L
),
so
that
3
+
−
+
(S3− ∩ Q+
2 ) ∪ V (L ) ∪ (Q3 ∩ S2 )
−
lies in the intersection (T1+ \ Q−
4 ) ∩ T1 , and contains a path joining the points
assigned to the triples from (6.4).
To see that this L+ is indeed the edge-set of a path, suppose it is false. In
particular, we can, in cyclic order in ∂Q−
4 , find edges e1 , e2 , e3 , e4 such that
e1 , e3 ∈ L+ and e2 , e4 ∈ L− \ L+ . But then since T1+ \ Q−
4 is connected, there
is a path of squares in T1+ \ Q−
from
e
to
e
.
But,
together
with the squares
1
3
4
of Q−
,
these
squares
must
enclose
either
s
or
s
,
where
s
is
the
square incident
2
4
i
4
−
+
−
with ei lying outside of Q4 . In particular, T1 \ Q4 cannot be simply connected,
contradicting the inductive hypothesis (T1).
Thus the hypotheses of Lemma 5.1 are satisfied for (T̃ , Ẽ). In particular, the
tiles in T̃ form a tiling, the only intersecting pairs of tiles in T̃ are those in E, and
all nonempty intersections of tile-pairs are paths.
We are now ready to verify (T1)-(T5) for the quadruple (C0 , C1 , C2 , C3 ).
(T1) T0 is a tile, and |T0 | = c0 . Moreover, T0 \ Ti± is a tile, which touches Ti± , for
each i = 1, 2, 3.
Proof. Since T0 is a set of squares whose centers form a connected subgraph of the
∗
dual lattice Z2 , the fact that T0 is a tile follows from the fact that T0 + L covers
the plane without overlap (so T0 has no “holes”), by Lemma 5.1. Similarly, since
any 5 tiles from among
−
+
−
+
−
T1+ \ Q−
4 , T1 , T2 , T2 , T3 , T3
induce a connected subgraph of (T̃ , Ẽ) whose complement is also connected, we get
that T0 \ Ti± is a tile. It also follows that |T0 | = c0 , since c0 is the determinant of
L.
(T2) T0 is 90◦ symmetric.
Proof. Claim 6.5 implies that the tiles Rik , for i = 1, 2, 3 and k = 1, 2, 3, 4, surround
T0 . By induction, each of the Rik is 90◦ symmetric. Moreover, using the lattice
rules (4.1), we see that the union S = ∪Rik is a 90◦ symmetric union of squares.
INTEGER SUPERHARMONIC MATRICES
31
Since T0 is the bounded component of the complement of S \ ∂S, it too must be
90◦ symmetric.
(T3) T0 , T0 + vi0 , and T0 − vj0 form a touching triple, provided c0 > 1.
Proof. Our use of Lemma 5.1 tells us that the only pairs of tiles in T̃ which intersect
are those in Ẽ. Thus the pairwise intersections of the tiles T0 , T0 + v0i , and T0 − v0j
can be written as the union of the subtiles from the respective tiles which are drawn
adjacently in Figure 6.3. Each of these intersections is a simple path between points
ρ(F ), ρ(F 0 ) which can be concatenated to a simple path. (The resulting path is
simple since the subtiles are all nonoverlapping.) We are done by (T2).
(T4) T0 , T0 + vi0 , and Ti form a touching triple, if c(Ti ) − c(T0 ) = 21 (vi0 + v0i ),
provided c0 > 1.
Proof. Using the lattice rules (4.1), one can verify for each i in (1, 2, 3) that Ti = Ri1 ,
as defined in (6.3). In particular, as above, Lemma 5.1 implies that T0 , T0 + v0i ,
and Ri1 form a touching triple. The statement follows now from (T2).
(T5) T0 , Ti , and Tj form a touching triple, if c(Ti ) − c(T0 ) =
c(Tj ) − c(T0 ) = 12 (vj0 − v0j ).
1
2 (vi0
+ v0i ), and
Proof. Using the lattice rules (4.1), we can verify that Tj = Rj2 Lemma 5.1 gives
that T0 , Ri1 , and Rj2 form a touching triple. Thus, since we also had that Ti = Ri1 ,
the statement follows now from (T2).
(T6) If c1 ≥ c2 > 1, then among other labeled tiles from Figure 6.3, the subtile
Ti± intersects only those which are drawn adjacent to it or with overlap. Moreover,
+
+
−
+
−
F (T1+ ) ∩ F (T2+ ) ⊆ F (Q+
2 ) ∩ F (T2 ), and F (T1 ) ∩ F (T1 ) ⊆ (F (Q2 ) ∩ F (S3 )) ∪
−
+
−
F (Q4 ) ∪ (F (Q3 ) ∩ F (S2 )).
Proof. These are both consequences of the application of Lemma 5.1. The first
follows from the fact that F (T1 ) ∩ F (T2 ) is a path between ρ(F ) and ρ(F 0 ) for the
faces F, F 0 which contain T1 , T2 . Similarly, the second follows from the fact that
−
+
−
−
+
−
F (T1+ \Q−
4 )∩F (T1 ) is a path in (F (Q2 )∩F (S3 ))∪F (Q4 )∪(F (Q3 )∩F (S2 )). 7. Boundary Strings
Having constructed tiles, we now turn our attention towards constructing odometers. The main idea is to mimic the tile construction, using the duality between the
rules for the vij and aij in (4.1) to attach superharmonic data to the tiles. However,
there is a problem in directly lifting the tile construction: the odometers are only
180◦ symmetric in general, and we used 90◦ symmetry in constructing the tiles.
A examination of the argument in Section 6 suggests that we made essential use
of 90◦ symmetry only to conclude that the lattice ΛC generated by {v10 , v20 , v30 } is
a tiling lattice for T0 from the fact that iΛC is a tiling lattice for T0 . To give a proof
using only 180◦ symmetry, we must therefore find a way to express the interface
between T0 and T0 + vi0 directly in terms of relationships between subtiles.
We need a new type of tile decomposition, which we call boundary strings. If T
is a regular tiling, then we call a sequence T 0 , T 1 , . . . , T n of tiles in T the T -string
from T 0 to T n if:
(1) T i touches T i+1 for 0 ≤ i < n,
32
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
T0
T
1
T3
T5
T7
T2
T4
T6
Figure 7.1. A string T 0 , . . . , T 7 in the regular tiling TC + ΛC for
the circle C = (28, 7 + 20i).
(2) Each c(T i ) (0 < i < n) lies in the closed half-plane to the left of the ray
from c(T 0 ) to c(T n ); i.e.:
c(T i ) − c(T 0 )
Im
≥0
c(T n ) − c(T 0 )
(3) Each T i (0 < i < n) touches some S i ∈ T whose centroid lies outside of
the closed half-plane to the left of the ray from c(T 0 ) to c(T n ).
A string is the left-handed approximation of the line segment from c(T 0 ) to c(T n )
in T ; see Figure 7.1. Observe that there is a unique string between any two tiles
in a regular tiling. The interior tiles of the L-string S from T 0 to T n are the tiles
in the string other than the endpoints T 0 , T n , and the interior of the string is the
union of the footprints of all interior tiles.
Given a tile T0 for the circle C0 in the proper Descartes quadruple (C0 , C1 , C2 , C3 ),
the Ci boundary-string (i = 1, 2, 3) for the tile T0 is the string from the tiles Ri− to
Ri+ for Ci , where
c(Ri± ) = c(T0 ) + 21 (vi0 ± v0i ).
The following lemma shows that the boundary strings for a tile can be constructed by concatenating certain smaller strings together.
Lemma 7.1. Let (C0 , C1 , C2 , C3 ) ∈ B 4 be a proper Descartes quadruple, write
vij = v(Ci , Cj ), and suppose T0 is a tile for C0 with the tile decomposition {Ti± },
and suppose i ∈ {1, 2, 3}, ci > 0, and that Ri± is a tile for Ci satisfying
c(Ri± ) = c(T0 ) + 21 (vi0 ± v0i ).
Then we have that
c(Ri+ ) − c(Ti− ) = vki ,
c(Ti− ) − c(Ri− ) = −vji ,
(7.1)
and the (Ri− + ΛCi )-string from Ri− to Ri+ is the concatenation of the (Ri− + ΛCi )strings from Ri− to Ti− and from Ti− to Ri+ , respectively.
Note that (7.1) implies via Lemma 4.3 that the tiles Ti− and Ri+ both lie in tiling
+ ΛCi , ensuring that referenced strings are well defined.
Ri−
INTEGER SUPERHARMONIC MATRICES
33
T30
T35
T33
T34
T31
T23
T32
T22
T10
T21
T11
T20
T12
Figure 7.2. The three boundary strings of a tile.
It is not hard at this point to use the lattice rules to strengthen Lemma 7.1, to
show inductively that the strings from Ri− to Ti− and from Ti− to Ri+ are themselves
boundary strings of subtiles (when they have more than two tiles). This induction
also gives, for example, that the interior tiles of a boundary string for T0 lie in T0 ,
as seen in Figure 7.2. We postpone this calculation until the next section, however,
when we are prepared to simultaneously show that the strings have important
compatibility properties with respect to our odometer construction.
Proof of Lemma 7.1. The offsets in (7.1) result from a straightforward calculation
using (4.1); for convenience, note that, referring to the same same tile collection
(6.3) used in the proof of Lemma 6.3 (see Figure 6.3) one can check that Ri+ = Ri1 ,
Ri− = Ri0 in that decomposition.
Examining the definition of a string, we see that the statement regarding the
concatenation fails only if the interior of the triangle 4c(Ti− )c(Ri− )c(Ri+ ) contains
the center of some tile in the tiling Ti− + ΛCi . However, the lattice generated by
−vji and vki has determinant
1
(vik vji + vik vji ) = ci
2
by the lattice rules (4.1). In particular, the triangle 4c(Ti− )c(Ri− )c(Ri+ ) has area
1
2 ci . Thus the lemma follows from the fact that any triangle of area half the
determinant of a lattice containing its three vertices can contain no other points of
the lattice. (This is a special case of Pick’s theorem, for example.)
The following topological lemma allows us to analyze tile interfaces – and, in
particular, odometer interfaces – using only 180◦ symmetry.
Lemma 7.2. If R and S are tiles in the tiling T = T0 + ΛC0 corresponding to the
circle C0 ∈ B with c0 ≥ 1, then the intersection of the interiors of the T -string R
from R to S and the T -string S from S to R contains a path in Z[i] from R to S.
Proof. We consider the graph G on T where T, T 0 are adjacent if they touch. G is
isomorphic to the graph of the triangular lattice unless c0 = 1, in which case the
lemma is easy to verify directly.
34
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
For the former case, we draw G in the plane by placing each vertex at the center
of the corresponding tile, and drawing straight line segments between adjacent
vertices of G; the result is a planar embedding of G which is affine-equivalent to
the equilateral triangle embedding of the triangular lattice.
We now consider the sequences F 0 , F 1 , . . . , F k of faces of G through which the
line segment from c(R) to c(S) passes, in order. If we associate to each F the point
ρ(F ) in Z[i] lying in the intersection of the three tiles of F , then the consecutive faces
F i and F i+1 either share an edge corresponding to touching tiles T ∈ R, T 0 ∈ S, or
they share a vertex corresponding to a tile in both R and S; in either case, there
is a path in Z[i] from ρ(F ) to ρ(F 0 ) lying in the intersection of the interiors of S
and R. Concatenating these paths consecutively, we get a walk in Z[i] from ρ(F 0 )
to ρ(F k ), lying in the intersection of the interiors of R and S.
8. Tile odometers
In this section we attach function data to our tiles. Since we are no longer
concerned with topological issues, our definition of a tile as a set of squares sx is
no longer useful, and from here on by a tile T for a circle C we mean the footprint
of the tile constructed for C in Section 6. In particular, T ∩ T 0 is now denotes a
subset of Z2 .
A partial odometer is a function h : T → Z with a finite domain T (h) ⊆ Z2 . We
write s(h) ∈ C for the slope of h, which is the average of
1
(h(x + 1) − h(x) + h(x + 1 + i) − h(x + i)) +
2
i
(h(x + i) − h(x) + h(x + 1 + i) − h(x + 1)) (8.1)
2
over squares {x, x + 1, x + i, x + 1 + i} ⊆ T ; this is a measure of an average gradient
for h. Note that the slope is not defined when T is a singleton. We say that two
partial odometers h1 and h2 are translations of one another if
T (h1 ) = T (h2 ) + v
and h1 (x) = h2 (x + v) + a · x + b,
(8.2)
2
for some v, a ∈ Z and b ∈ Z.
Definition 8.1. We say that two partial odometers h1 and h2 are compatible if
h2 − h1 = c on T (h1 ) ∩ T (h2 ) for some constant c, which we call the offset constant
for the pair (h1 , h2 ), or if T (h1 ) ∩ T (h2 ) = ∅.
When the offset constant is 0, or in the second case, we write h1 ∪h2 for the common
extension of the hi to the union of their domains. The next lemma which allows us
to glue together pairwise compatible partial odometers. Recall we have defined a
tiling as a collection of tiles T such that every square sx = {x, x + 1, x + i, x + 1 + i}
of Z2 is contained in exactly one element of T .
Lemma 8.2. If H is a collection of pairwise compatible partial odometers such that
T = {T (h) : h ∈ H} is a hexagonal tiling, then there is a function g : Z2 → Z,
unique up to adding a constant, which is compatible with every h ∈ H.
Proof. Since T is a hexagonal tiling, its intersection graph G is a planar triangulation, and each face {h0 , h1 , h2 } of G corresponds to a touching triple of tiles T (hi )
with T (h0 ) ∩ T (h1 ) ∩ T (h2 ) 6= ∅. In particular, writing d(hi , hj ) for the constant
value of hi − hj on T (hi ) ∩ T (hj ), we have d(h0 , h1 ) + d(h1 , h2 ) + d(h2 , h1 ) = 0.
INTEGER SUPERHARMONIC MATRICES
35
Thus d is an edge function with zero curl, so it can be written as the gradient of
a vertex function f which is unique up to additive constant. Now fix any h0 ∈ H
and for x ∈ T (h) set g = h + f (T (h)) on T (h).
Our goal is to associate a partial odometer h, unique up to odometer translation
(8.2), to every circle C ∈ B.
Definition 8.3. A partial odometer h0 : T0 → Z is a tile odometer for C0 ∈ B if
T0 is a tile for C0 and either:
• C0 = (0, ±1) (so T0 is a singleton),
• C0 = (1, 1 + 2z) for some z ∈ Z[i], and h is any translation of the partial odometer h0 : {0, 1, i, 1 + i} → Z given by h(0) = h(1) = h(i) = 0,
h(1 + i) = Im(z)/2, or
• (C0 , C1 , C2 , C3 ) ∈ B 4 is a proper Descartes quadruple, Ci = (ci , zi ), aij =
a(Ci , Cj ), vij = v(Ci , Cj ), and
−
+
−
+
−
h0 = h+
1 ∪ h1 ∪ h2 ∪ h2 ∪ h3 ∪ h3 ,
where
h±
i
(8.3)
is a tile odometer for Ci such that
1
c(T (h±
i )) − c(T (h0 )) = ± 2 (vkj − ivkj ),
and
1
ci = 0 or s(h±
i ) − s(h0 ) = ± 2 (akj + iakj ),
for all rotations (i, j, k) of (1, 2, 3).
We call the h±
i ’s the subodometers of h0 .
When T1 , ..., Tn is a sequence of tiles such Tk is a subtile of Tk+1 , we call T1
an ancestor tile of Tn . Restrictions h|T of odometers to ancestor tiles T of T (h)
are called ancestor odometers of h0 . The following lemma asserts that this makes
sense:
Lemma 8.4. If h0 : T0 → Z is a tile odometer for C0 and the tile T for the circle
C ∈ B is an ancestor of T0 , then h0 |T is a tile odometer for C.
By induction, tile odometers are easily seen to be centrally symmetric:
Lemma 8.5. The reflection x 7→ h0 (−x) on the tile −T0 is a translation of h0 .
Our main goal in this section is to prove inductively that each circle C ∈ B has
an associated tile odometer. (Note that, inductively, the definition immediately
gives that tile odometers for a given circle are unique up to odometer translation.)
However, before proving that circles in B do have tile odometers, we will prove that
tile odometers must have certain compatibility properties when they do exist.
Given a proper Descartes quadruple (C0 , C1 , C2 , C3 ) with c0 > 0, we say that
tile odometers h0 and h00 for C0 are left-lattice adjacent if
c(T (h00 )) − c(T (h0 )) = ±vi0
s(h00 ) − s(h0 ) = ±ai0
for some i ∈ {1, 2, 3} (with matching signs, as usual) and similarly right-lattice
adjacent if
c(T (h00 )) − c(T (h0 )) = ±v0i
s(h00 ) − s(h0 ) = ±a0i .
36
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
for some i ∈ {1, 2, 3}. Given tile odometers h0 and hi for C0 and Ci (i = 1, 2, 3),
we say that hi is subtile-lattice adjacent to h0 if
c(T (hi )) − c(T (h0 )) = is 21 (vi0 + v0i )
s(hi ) − s(h0 ) = (−i)s 21 (ai0 + a0i ).
for some s ∈ {0, 1, 2, 3}. Note that the subtile-lattice adjacency relationship is not
symmetric.
Note that the cases of left-lattice adjacency and subtile-lattice adjacency, respectively, correspond for the domains of h0 and h0t (t = 0, 1, 2, 3) to the two types of
pairwise tile relationships which were seen in (T4). Our next goal is to show that
if two tile odometers are left-lattice adjacent or right-lattice adjacent then they are
compatible in the sense of Definition 8.1. The following lemma will allow induction
from the right-lattice and subtile-lattice adjacent cases.
Lemma 8.6. If the tile odometers h and h0 are right-lattice adjacent, then any
subodometer of h intersecting h0 is subtile-lattice adjacent to h0 . Similarly, if the
tile odometer h is subtile-lattice adjacent to h0 , then for any subodometer h±
i of
h0 which intersects h, either h, h±
are
left-lattice
adjacent,
or
h
is
subtile-lattice
i
±
adjacent to h±
i , or hi is subtile-lattice adjacent to h.
Proof. Recall the tiles Rij defined (6.3) from the proof of Lemma 6.3, as shown in
Figure 6.3; each Rij is defined by
Ri0 = Ti+ + v0k
Ri1 = Ti+ − v0j
Ri2 = Ti− − v0k
Ri3 = Ti− − v0j ,
where (i, j, k) is a rotation of (1, 2, 3). We define for each Rit a translation hti of the
tile odometer for the circle Ci whose domain is Rij , by
c(h0i ) = c(h+
i ) + v0k
s(h0i ) = s(h+
i ) + a0k
c(h1i ) = c(h+
i ) − v0j
s(h1i ) = s(h+
i ) − a0j
c(h2i ) = c(h−
i ) − v0k
s(h2i ) = s(h−
i ) − a0k
c(h3i ) = c(h−
i ) + v0j
s(h3i ) = s(h−
i ) + a0j
We know from (T4) and the application in Section 6 of Lemma 5.1 that the boundary of T0 is covered by the Rij ’s. Thus, to prove the first part of the lemma, it is
sufficient to show compatibility of h0 with the hti . And by part (T6) of Lemma 6.3,
t
it is sufficient to show compatibility of tile odometers h±
i and hs whose domains are
drawn as touching in Figure 6.3. It can be checked by hand using the lattice rules
t
(4.1) that such a h±
i and hs are either left-lattice adjacent or subtile-lattice adjacent, proving the full statement of the lemma, since the hts include all odometers
which are subtile-lattice adjacent to h0 .
Looking ahead, if we knew left-lattice adjacent tile odometers to be compatible,
then by induction, Lemma 8.6 would give compatibility of right-lattice adjacent
and subtile-lattice adjacent odometers as well. Indeed, we give this as Lemma 8.9,
below.
INTEGER SUPERHARMONIC MATRICES
37
−
R3
T3−
+
R3
T1+
T2+
Figure 8.1. Verifying a boundary string.
Unlike the argument for tiles, we can not apply 90◦ symmetry and v0i = ivi0 to
add the case of left-lattice adjacent tile odometers to Lemma 8.6, since odometers
are only 180◦ symmetric in general. Instead, we will express the shared boundary
of the touching tiles in terms of boundary strings, and use the compatibility of the
restrictions of the odometers to the tiles making up the boundary strings.
To do this, we first need to strengthen our notion of boundary string: We say a
partial odometer respects a string when its domain includes all tiles of the string,
and its restrictions to those tiles are tile odometers which are consecutively leftlattice adjacent.
Lemma 8.7. Suppose (C0 , C1 , C2 , C3 ) is a proper Descartes quadruple, h0 is a tile
odometer for the circle C0 with domain T0 , and write vij = v(Ci , Cj ). For each
i = 1, 2, 3 for which ci > 0, we have that if Ri± are the endpoints of the Ci boundary
string R for T0 and each hR± is a tile odometer for Ci with domain Ri± , which
i
satisfies
s(hR± ) − s(h0 ) = 21 (vi0 ± v0i ),
i
then hR+ hR− , and h0 are compatible, and h0 ∪ hR+ ∪ hR− respects the Ci boundary
i
i
i
i
string of T0 .
Proof. Recall from the definition of boundary strings that
c(Ri± ) − c(T0 ) = 21 (vi0 ± v0i ).
Thus each hR± is subtile-lattice adjacent to h0 , and thus compatible with h0 by
i
Lemma 8.9. Moreover, Lemma 7.1 gives
c(Ri+ ) − c(Ti− ) = vki
and c(Ti− ) − c(Ri− ) = −vji .
(8.4)
R1− , Ti− , R1+
In particular, for i = 1, then our assumption that c1 > c2 , c3 implies that
are a triple of consecutively touching tiles in a tiling of T1− under the lattice generated by {v41 , v21 , v31 }; this is already sufficient to imply that they form a string
(either from R1− to R1+ or vice versa), and the sign in (4.1e) implies that they are
the string from R1− to R1+ . Moreover, since
1
s(h−
1 ) − s(h0 ) = − (a32 + ia32 ),
2
we can calculate using the lattice rules (4.1) that
s(hR+ ) − s(h−
i ) = a31 ,
1
s(h−
1 ) − s(hR− ) = −a21 ,
1
(8.5)
38
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
T0 + vi0
Si2
Si1
Si0
Ri3
Ri4
Si4
Si3
Ri0
Ri1
Ri2
T0
Figure 8.2. Verifying tiling TC0 by ΛC0 using boundary strings.
and (8.4) and (8.5) together give that hR− , hT − , hR+ are consecutively left-lattice
1
i
i
adjacent; thus h0 ∪ hR+ ∪ hR− respects the string.
i
i
The cases i = 2, 3 require the induction hypothesis and, since they are similar,
we handle only the case i = 3. Decomposing h0 according to tile odometers h±
i on
±
tiles Ti according to Definition 8.3 and using the lattice rules (4.1), we check that
c(R3+ ) − c(T2+ ) = 12 (v32 − v23 )
and c(T3− ) − c(T2+ ) = 21 (v32 + v23 ),
1
s(hR+ ) − s(h+
2 ) = 2 (a32 − a23 )
+
1
and s(h−
3 ) − s(h2 ) = 2 (a32 + a23 ),
3
(see Figure 8.1). In particular, the string S 0 from T3− to R3+ is the C3 boundary
string of T2+ , and by induction, h0 ∪ hR+ ∪ hR− respects this string. In exactly the
i
i
same way, we can check that the string S from R3− to T3− is the C3 boundary string
of T1+ , and is respected by h0 ∪ hR+ ∪ hR− . By Lemma 7.1, the C3 -string for T0
i
i
is the concatenation of S and S 0 , and this string is respected by h0 ∪ hR+ ∪ hR−
i
i
since all pairs of consecutive tiles in the concatenation are already consecutive tiles
in either S or S 0 .
We now prove the compatibility of left-lattice adjacent odometers:
Lemma 8.8. If the odometer h00 for C0 is left-lattice adjacent to h0 , then h00 and
h0 are compatible. Moreover, if ci > 1, where c(T (h00 )) − c(T (h0 )) = ±vi0 , then
any vertex in T (h0 ) ∩ T (h00 ) lies, together with all of its lattice neighbors, in the
union of the domains of some pair of proper ancestor odometers of h0 and h00 which
are pairwise left-lattice adjacent.
Proof. If ci = 0 then C0 is a Ford circle and this lemma is easy to verify from the
construction in Section 3.2, so we may assume ci > 0.
From the definition of left-lattice adjacency, we have without loss of generality
that h00 = hi0 (x) = h0 (x − vi0 ) + ai0 · x of h0 with domain T0i = T0 + vi0 , and we let
INTEGER SUPERHARMONIC MATRICES
39
Ri and Si be tiles for Ci satisfying
c(Ri ) − c(T0 ) = 21 (vi0 − v0i ),
c(Si ) − c(T0 ) = 21 (vi0 + v0i ).
From Lemma 6.3, we know that Ri− , T0 , T0i and Ri+ , T0 , T0i are both touching
triples; in particular, the intersection of T0 and T0i is a path from Si to Ri . Therefore, letting R = Ri0 , Ri1 , . . . , Rit and S = Si0 , Si1 , . . . , Sit be the string from Ri− = Ri0
to Ri+ = Rit and from Ri+ = Si0 to Ri− = Sit , respectively (Figure 8.2), Lemma 8.7
guarantees that the interiors of R and S lie in T0 and T0 + vi0 , respectively, and
Lemma 7.2 implies that their intersection contains a simple path from from Si to
Ri , we have that T0 ∩ T0i lies in the intersection of the interiors of R and S. So it
suffices to show that each restriction for each pair of touching Ri`1 , Si`2 from R and
S that h0 |Rit and h0 |Si` are left-lattice adjacent (thus compatible by induction).
Note that ci > 1 implies that there is no vertex in Z2 which lies in the intersection
of 4 tiles in a Ti + ΛCi tiling of Z2 , justifying the Moreover clause.
We let hR± be defined as in Lemma 8.7, and let f0i = h0 ∪ hR− ∪ hR+ and
i
i
i
i
f1 = hi0 ∪ hR− ∪ hR+ , which, by Lemma 8.7, are well-defined partial odometers
i
i
which respect the strings R and S, respectively. Considering now touching tiles
Ri`1 and Si`2 from R and S, respectively, we see that
Ri`1 , Ri`1 −1 , . . . , Ri0 = Sit , Sit−1 , . . . , Si`2
is a sequence of tiles where each consecutive pair U m , U m+1 in the sequence forms a
two-tile string which is respected either by f0i or g0i . In particular, letting Ci1 , Ci2 , Ci3
be the parents of Ci in clockwise order, we have each pair U m , U m+1 respected by
fqi (q ∈ {0, 1}) satisfies
s(fqi |U m+1 ) − s(fqi |U m ) = ±a(Ci , Cin ),
c(U
m+1
m
) − c(U ) =
where
±v(Ci , Cin ).
(8.6)
(8.7)
We now have
±v(Ci , Cin ) = c(Si`2 ) − c(Ri`1 ) =
X
c(U m+1 ) − c(U m )
implies that
s(f1i |Si`2 ) − s(f0i |Ri`1 ) =
X
s(fqim |U m+1 ) − s(fqim |U m ) = ±a(Ci , Cin ).
In particular, f1i |Si`2 and f0i |Ri`1 are left-lattice adjacent.
As noted earlier, Lemma 8.6 now gives us the following:
Lemma 8.9. If h is a tile odometer which is right-lattice or subtile-lattice adjacent
to the tile odometer h0 , then h and h0 are compatible.
Finally, we prove the existence of tile odometers for each circle in B.
Lemma 8.10. Every circle C0 ∈ B has a tile odometer h0 : T0 → Z.
Proof. In light of Section 3, we may assume that (C0 , C1 , C2 , C3 ) is a proper
Descartes quadruple with c1 ≥ c2 > 1. Copying the proof of Lemma 6.2, we
easily obtain tile odometers h±
i for each circle Ci such that
1
c(T (h±
i )) − c(T0 ) = ± 2 (vkj − ivkj ),
40
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
and
ci = 0
1
or s(h±
i ) − s(h0 ) = ± 2 (akj + iakj ),
for all rotations (i, j, k) of (1, 2, 3). The difficulty lies in checking there are height
offsets so that the h±
i have a common extension to T0 .
We know from part (T6) of Lemma 6.3 that the only pairs of subodometers h±
i
whose domains intersect are the pairs
∓
(h±
i , hj )
for i 6= j,
±
(h±
1 , h2 ),
−
(h+
1 , h1 ).
and
Thus, as in the proof of Lemma 8.2, we only need to check compatibility of these
∓
pairs. The tile odometers h±
i , hj for i 6= j are easily verified to be subtile-lattice
adjacent, and so are compatible by Lemma 8.9. It remains to check compatibility
for tile odometer pairs corresponding to the tile pairs (T1− , T2− ), (T1+ , T2+ ), and
(T1+ , T1− ). To accomplish this, we decompose T1+ and T1− into unions of Q±
i and
Si± for i = 2, 3, 4 according to (6.1), as shown in Figure 6.2. The second part of (T6)
implies that it suffices to check compatibility of the restrictions of the odometers
−
−
+
+
+
−
+
−
h±
i to the pairs (S2 , T2 ), (Q2 , T2 ), (Q2 , S3 ), and (Q3 , S2 ). Using the lattice
rules we verify that these pairs are right-lattice, right-lattice, subtile-lattice, and
subtile-lattice adjacent, respectively, so we are done by induction.
9. Global odometers
Having constructed tile odometers hC for each circle C ∈ B by Lemma 8.10,
we now check that the hC extend to global odometers gC satisfying ∆gC ≤ 1, and
prove our main theorem. We first observe that the partial odometers glue together
to form global odometers with the correct periodicity and growth at infinity. (As
in the previous section, a tile now is just a subset of Z[i].)
Lemma 9.1. For every circle C ∈ B, there is a function gC : Z2 → Z which has a
restriction to a tile odometer for C, for which the periodicity condition (1.5) holds
for v ∈ ΛC , and for which
x 7→ gC (x) − 21 xt AC x − b · x,
(9.1)
is ΛC -periodic for some b ∈ R2 .
Proof. We may suppose C = C0 and (C0 , C1 , C2 , C3 ) is a proper Descartes quadruple, hi is a tile odometer for Ci whose domain is the tile Ti , and write vij =
v(Ci , Cj ), aij = a(Ci , Cj ). Pairs of tile odometers with overlapping domains from
the collection
H = {x 7→ h(x − k1 v10 − k2 v20 ) + (k1 a10 + k2 a20 ) · x : k1 , k2 ∈ Z},
are left-lattice adjacent, and thus compatible. By Lemma 8.2, there is a function
g : Z2 → Z with g(0) = 0 that is compatible with every h ∈ H. Since H is invariant
under
h 7→ (x 7→ h(x − vi0 ) + ai0 · x)
for each i ∈ {1, 2, 3}, we see from Lemma 8.2 that
x 7→ g(x − vi0 ) + ai0 · x,
INTEGER SUPERHARMONIC MATRICES
41
differs from g by some constant:
g(x + vi0 ) = βi + ai0 · x + g(x).
But g(0) = 0 implies that g(vi0 ) = βi , so that
g(x + vi0 ) = g(vi0 ) + ai0 · x + g(x),
for all i = 1, 2, 3 and x ∈ Z2 . Together with ai0 = AC0 vi0 from (4.2), this implies
the periodicity condition (1.5) for v ∈ ΛC and that (9.1) is ΛC -periodic for some
b ∈ R2 .
To prove this criterion for general odometers gC , we follow the outline of Proposition 3.3. To begin, we need to understand the Laplacian ∆gC for all C ∈ B. Let
N (x) and N̄ (x) denote the set {x ± 1, x ± i} of lattice neighbors of x ∈ Z[i] and
{x} ∪ N (x), respectively.
Lemma 9.2. Let h1 and h2 be compatible tile odometers for (tangent or identical)
circles in B, let h = h1 ∪ h2 , and let x such that x ∈ T (h1 ) ∩ T (h2 ) and N̄ (x) ⊆
T (h1 )∪T (h2 ). If h1 , h2 are left-lattice adjacent, then ∆h(x) = 1. If h1 , h2 are rightlattice adjacent or subtile-lattice adjacent, then ∆h(x) = 0 if x ∈ sy 6⊆ T (h1 )∪T (h2 )
for some y, and ∆h(x) = 1 otherwise.
Proof. The proof is by induction on the areas of T (h1 ) and T (h2 ). For the cases of
right-lattice and subtile-lattice adjacency, the base case occurs when h1 and h2 are
both tile odometers for a circle of curvature 1, in which case the statement can be
checked by hand; it is sufficient to check for the circle (1, 1). Thus for the inductive
step in this case, we assume (without loss of generality) h1 is a tile odometer for
a circle of curvature > 1: in particular, it can be decomposed into subodometers
according to Definition 8.3.
For this case we let x be a point such that N̄ (x) ⊆ T (h1 ) ∪ T (h2 ). We claim
that there is a subodometer h0 of h1 such that N̄ (x) ⊆ T (h0 ) ∪ T (h2 ). Indeed,
the definition of a tile ensures that if hi (i = 1, 2) covers sx and sx−1−i then
it must also cover either sx−1 or sx−i . (In particular, at least three of the four
squares containing x as a vertex are covered by h.) Thus, without loss of generality,
we have that sx is covered by h2 and sx−1−i is covered by h1 . In this case we
let h0 be the subodometer of h1 whose domain covers sx−1−i , and we have that
N̄ (x) ⊆ T (h0 ) ∪ T (h2 ). Lemma 8.6 now implies that h0 and h2 are subtile-lattice
adjacent or left-lattice adjacent, so we are done by induction (in particular, note
that the condition on sy which determines whether ∆h(x) is 0 or 1 is unchanged
with the inductive step).
For the case of left-lattice adjacency, Lemma 8.8 gives the statement by induction. The base case is when h1 and h2 are left-lattice adjacent along a vi0 for which
the curvature ci of the corresponding parent circle Ci is 1. In this case, however,
the proof of Lemma 8.8 gives that x lies, together with all of its neighbors, in the
union of four tile odometers for Ci = (1, 1 + 2z) (z ∈ Z[i]) which are cyclically
left-lattice adjacent, and this case can be checked by hand.
Note that from the case of left-lattice adjacency, Lemma 9.2 has the crucial consequence that ∆gC ≡ 1 on the “web” of its tile boundaries, as seen in Figure 1.2.
Next, we analyze ∆gC on the interior of TC .
Lemma 9.3. Suppose (C0 , C1 , C2 , C3 ) ∈ B 4 is a proper Descartes quadruple, write
Ci = (ci , zi ) and Ti is a tile for Ci , and decompose T0 as a union of Ti± as in (6.1).
42
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
Let x ∈ T0 \ ∂T0 , and let k denotes the number of boundaries ∂Ti± of subtiles that
contain x. Then we have
(1)
(2)
(3)
(4)
(5)
If
If
If
If
If
k
k
k
k
k
=2
=2
=3
=4
=4
and x 6= c(T0 ) then
and x = c(T0 ) then
then ∆gC0 (x) = 0.
and x 6= c(T0 ) then
and x = c(T0 ) then
∆gC0 (x) = 1.
∆gC0 (x) = 0.
∆gC0 (x) = −1.
∆gC0 (x) = −2.
In particular, as 180 degree symmetry precludes k = 3 when x = c(T0 ), we have
∆gC0 (x) = 3 − k − 1{c(T )} (x)
whenever k(x) ≥ 2. Note that cases (4) and (5) arise only when C0 is a Ford or
Diamond circle, so these cases have been proved already in Section 3.
The above lemma provides a fast algorithm for recursively generating tiles with
their associated Laplacian patterns; the base case for the recursion is given by the
base cases of Definition 8.3. The appendix lists all such patterns associated to
circles in B of curvature 1 ≤ c ≤ 100.
Proof of Lemma 9.3. For the Ford and Diamond circles, this Lemma has already
been verified in Section 3.2. Thus, we may assume by induction that the ci are
distinct and positive, and that the lemma holds for the proper Descartes quadruples
(C1 , C4 , C2 , C3 ) and (C2 , C3 , C5 , C6 ) for C4 , C5 , C6 ∈ B. Since c1 > 1, we can
±
decompose T1+ and T1− into Q±
i and Si as in Claim 6.6 and Figure 6.2.
First consider the case where k = 3. Definition 8.3 (and part (T6) from Lemma 6.3)
give that the two subtiles containing x are subtile-lattice adjacent. Since k = 3,
there exists the square sy in the final hypothesis of Lemma 9.2, and thus Lemma 9.2
gives that ∆gC0 (x) = 0.
Next consider the case where x 6= c(T0 ) and k = 2. If the two subtiles whose
boundaries contain x are not the pair {T1+ , T1− }, then by Definition 8.3 (and part
(T6) from Lemma 6.3), the two subtiles containing x are subtile-lattice adjacent,
and Lemma 9.2 now gives the result.
If on the other hand k = 2 and x ∈ ∂T1+ ∩ ∂T1− , then we use the double
decomposition of T1+ , T1− . As in the proof of Lemma 9.2, we are guaranteed that
−
+
+
two tiles from among S3− , Q+
3 , S2 , Q2 , S4 cover the neighborhood N̄ (x). As in the
proof of Lemma 8.10, all pairs among these tiles are known to have induced tile
odometers which are subtile-lattice adjacent or right-lattice adjacent except for the
−
−
+
pairs {Q+
2 , S2 } and {S3 , Q3 }. But (T6) implies that this case cannot occur unless
ci = 0 for some i ∈ {3, 4}, in which case C0 is a Ford circle.
Finally, if x = c(T ), then x ∈ ∂T1+ ∩ ∂T1− implies that c4 = 0. In particular,
C1 = Cpq is a Ford circle with Ford circle parents C2 = Cp1 q1 and C3 = Cp2 q2 .
±
We have that N̄ (x) is covered by T1+ ∪ T1− . In this case, the tile odometers gpq
±
+
−
on T1 are related by qpq (x − (q2 − q1 , q)) = qpq (x) + (p, p2 − p1 ) · x + k for some
constant k ∈ Z. Thus, the explicit formula from Section 3.2 can be used to verify
this case.
We now generalize the inductive argument in Proposition 3.3 to obtain maximality of general odometers.
Lemma 9.4. For each C ∈ B, gC is maximal.
INTEGER SUPERHARMONIC MATRICES
43
Proof. Suppose X ⊆ Z2 is connected and infinite and ∆(gC + 1X ) ≤ 1. Let Y be a
connected component of Z2 \ X and observe that ∆(gC − 1Y ) ≤ 1. In particular, we
may assume that Y = Z2 \ X. It is enough to show that Y must be empty. Now,
if Y is not contained in T \ ∂T for some T ∈ TC + ΛC , then by Lemma 9.2, there
is a point x ∈ Y such that ∆gC (x) = 1 and ∆1X (x) > 0, since the tile odometers
of which gC consist are left-lattice adjacent. Thus, we may assume Y ⊆ TC \ ∂TC .
The lemma is now immediate from the following claim.
Claim. Suppose Y ⊆ TC is simply connected, Y \ ∂TC is non-empty, Y ∩ ∂T is
connected, and Y ∩ ∂TC ∩ Ti± is nonempty for at most one subtile Ti± . Then there
is a vertex x ∈ Y \ ∂TC such that ∆(gC − 1Y )(x) > 1.
We prove this by induction on the curvature of C. Note that, by the proofs
of Proposition 3.3 and Proposition 3.7, we may assume that C is neither a Ford
nor a Diamond circle. In particular, each Ti± contains at least one square and the
pairwise intersections are exactly what we expect from the picture. We may assume
that C1 is the largest parent. Thus Ti± and Tj∓ have simply connected intersection
when i 6= j and Ti± and Ti∓ are disjoint except when i = 1, in which case the
intersection is simply connected.
Case 1. Y is contained in the interior of some Ti± . The claim follows either by
induction hypothesis or the corresponding result for Ford and Diamond circles in
Proposition 3.3 and Proposition 3.7.
Case 2. Some x ∈ ∂Y \ ∂T lies in the boundary of exactly Ti+ and Tj− with
i 6= j. Observe that ∆1Y (x) < 0 and x 6= c(T ). Thus Lemma 9.3 gives the claim.
Case 3. Some x ∈ ∂Y \ ∂T lies in the boundary of exactly three Ti± . Since case
2 is excluded, we must have ∆1Y (x) < −1 and thus Lemma 9.3 again gives the
claim.
Case 4. In the exclusion of the above three cases, the topology of the tile
decomposition implies that Y lies in the union of the interior of T1+ , the interior of
T1− , and the intersection of one T1± ∩ ∂T . In Y ∩ T1+ is non-empty, then we can
inductively apply the claim to Y ∩ T1+ . Otherwise, we can inductively apply the
claim to Y ⊆ T1− .
From Lemma 9.1 and Lemma 9.4, we immediately obtain Theorem 1.2, modulo
checking that ΛC0 and LC0 are in fact the same lattice.
Theorem 9.5. LC = ΛC for all C ∈ B.
Proof. Lemma 4.2 verified ΛC ⊆ LC , thus it remains to verify LC ⊆ ΛC . For
C0 ∈ B, let gC0 be the odometer for C0 , as constructed in Section 9. Recall from
(1.5) that gC0 satisfies
gC (x + v) = gC (x) + xt AC v + gC (v)
for v ∈ ΛC0 . We will now modify gC0 to produce an odometer for C0 with periodicity
LC0 . In particular, for v ∈ LC0 , we let
v
gC
(x) = gC0 (x + v) − xt AC0 v − gC0 (v),
0
which is an integer since AC0 v ∈ Z2 by the definition of LC0 . Note that for v ∈ ΛC0
we have g v (x) = g(x).
We now define
g 0 (x) := min g v (x).
v∈LC0
44
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
Note that the periodicity g v (x) = g(x) for v ∈ ΛC0 implies that this can be interpreted as a finite minimum over the quotient LC0 /ΛC0 .
In particular, up to an additive constant, g 0 (x) is still an odometer for AC0 ,
and now satisfies the periodicity condition (1.5) for the lattice LC0 . In particular,
¯ 0 of g 0 over one period of LC must satisfy
we have that the average Laplacian ∆g
0
1
1
1
0
¯
¯
d = trace(AC0 ) ≤ ∆g ≤ ∆g = d . But then we must have det(LC0 ) ≥ d , and thus
LC0 = ΛC0 .
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
[13]
[14]
[15]
[16]
[17]
[18]
[19]
[20]
[21]
[22]
Per Bak, Chao Tang, and Kurt Wiesenfeld, Self-organized criticality: An explanation of the
1/f noise, Physical Review Letters 59 (1987), 381–384.
Jean Bourgain and Elena Fuchs, A proof of the positive density conjecture for integer Apollonian circle packings, Journal of the American Mathematical Society 24 (2011), no. 4, 945–967.
arXiv:1001.3894.
Jean Bourgain and Alex Kontorovich, On the local-global conjecture for integral Apollonian
gaskets, Inventiones mathematicae (2013), 1–62. arXiv:1205.4416.
Sergio Caracciolo, Guglielmo Paoletti, and Andrea Sportiello, Conservation laws for strings
in the Abelian Sandpile Model, Europhysics Letters 90 (2010), no. 6, 60003. arXiv:1002.3974.
John H. Conway, The sensual (quadratic) form, Carus Mathematical Monographs, vol. 26,
Mathematical Association of America, Washington, DC, 1997. With the assistance of Francis
Y. C. Fung.
Deepak Dhar, Self-organized critical state of sandpile automaton models, Physical Review
Letters 64 (1990), no. 14, 1613–1616.
Deepak Dhar, Tridib Sadhu, and Samarth Chandra, Pattern formation in growing sandpiles,
Europhysics Letters 85 (2009), no. 4, 48002. arXiv:0808.1732.
Deepak Dhar and Tridib Sadhu, A sandpile model for proportionate growth, Journal of Statistical Mechanics: Theory and Experiment 11 (2013), P11006. arXiv:1310.1359.
L. R. Ford, Fractions, American Mathematical Monthly 45 (1938), no. 9, 586–601.
Elena Fuchs, Counting problems in Apollonian packings, Bulletin of the American Mathematical Society 50 (2013), no. 2, 229–266.
Ronald L. Graham, Jeffrey C. Lagarias, Colin L. Mallows, Allan R. Wilks, and Catherine
H. Yan, Apollonian circle packings: geometry and group theory. I. The Apollonian group,
Discrete and Computational Geometry 34 (2005), no. 4, 547–585.
Alex Kontorovich and Hee Oh, Apollonian circle packings and closed horospheres on hyperbolic 3-manifolds, Journal of the American Mathematical Society 24 (2011), no. 3, 603–648.
arXiv:0811.2236.
Jeffrey C. Lagarias, Colin L. Mallows, and Allan R. Wilks, Beyond the Descartes circle
theorem, American Mathematical Monthly 109 (2002), no. 4, 338–361.
Lionel Levine, Wesley Pegden, and Charles K Smart, Apollonian structure in the Abelian
sandpile. Preprint (2012) arXiv:1208.4839.
Peter J. Nicholls, Diophantine approximation via the modular group, Journal of the London
Mathematical Society 17 (1978), no. 1, 11–17.
Srdjan Ostojic, Patterns formed by addition of grains to only one site of an abelian sandpile,
Physica A: Statistical Mechanics and its Applications 318 (2003), no. 1, 187–199.
Guglielmo Paoletti, Deterministic Abelian Sandpile Models and Patterns, 2012. Ph.D. thesis,
Department of Physics, Universita di Pisa. http://pcteserver.mi.infn.it/~caraccio/PhD/
Paoletti.pdf.
Wesley Pegden, Sandpile galleries. http://www.math.cmu.edu/~wes/sandgallery.html.
Wesley Pegden and Charles K Smart, Convergence of the Abelian Sandpile, Duke Mathematical Journal, to appear. arXiv:1105.0111.
Tridib Sadhu and Deepak Dhar, Pattern formation in growing sandpiles with multiple sources
or sinks, Journal of Statistical Physics 138 (2010), no. 4-5, 815–837. arXiv:0909.3192.
, The effect of noise on patterns formed by growing sandpiles, Journal of Statistical
Mechanics: Theory and Experiment 2011 (2011), no. 03, P03001. arXiv:1012.4809.
Peter Sarnak, Letter to J. Lagarias about integral Apollonian packings. http://web.math.
princeton.edu/sarnak/AppolonianPackings.pdf.
INTEGER SUPERHARMONIC MATRICES
[23] Katherine E. Stange,
arXiv:1208.4836.
The
sensual
Apollonian
circle
45
packing.
Preprint
(2012)
Appendix A. Table of Odometer Patterns
Here we display proper Descartes quadruples (C0 , C1 , C2 , C3 ) ∈ B, along with
the Soddy precursor C4 = 2(C1 + C2 + C3 ) − C0 of C0 , the vectors v(Ci , C0 )
and a(Ci , C0 ) (i = 1, 2, 3), the tile odometer for C0 , and a tiling neighborhood in
TC + LC . We display quadruples up to symmetry for 1 ≤ c0 ≤ 156. An extended
appendix with more circles is available as an ancillary file for this manuscript at
arXiv.org.
C0
C1
C2
C3
C4
v(C1 , C0 )
v(C2 , C0 )
v(C3 , C0 )
a(C1 , C0 )
a(C2 , C0 )
a(C3 , C0 )
(4, 1 + 4i)
(1, 1 + 2i)
(1, 1)
(0, −1)
(0, 1)
2 + 1i
−2 + 1i
−2i
1 + 1i
−1i
−1
(9, 1 + 6i)
(4, 1 + 4i)
(1, 1)
(0, −1)
(1, 1 + 2i)
3 + 2i
−3 + 1i
−3i
1 + 1i
−1i
−1
(12, 7 + 12i)
(4, 1 + 4i)
(1, 1 + 2i)
(1, 1)
(0, −1)
−4i
3 + 2i
−3 + 2i
−2 + 1i
2 + 1i
−2i
(16, 1 + 8i)
(9, 1 + 6i)
(1, 1)
(0, −1)
(4, 1 + 4i)
4 + 3i
−4 + 1i
−4i
1 + 1i
−1i
−1
(24, 17 + 24i)
(12, 7 + 12i)
(1, 1 + 2i)
(1, 1)
(4, 1 + 4i)
−6i
4 + 3i
−4 + 3i
−3 + 2i
3 + 1i
−3i
(25, 1 + 10i)
(16, 1 + 8i)
(1, 1)
(0, −1)
(9, 1 + 6i)
5 + 4i
−5 + 1i
−5i
1 + 1i
−1i
−1
(25, 1 + 20i)
(9, 1 + 6i)
(0, −1)
(4, 1 + 4i)
(1, 1)
−5 + 3i
−5i
5 + 2i
1 − 2i
−2
1 + 2i
tile odometer
One neighborhood of
TC in TC + LC .
46
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
(28, 7 + 20i)
(9, 1 + 6i)
(4, 1 + 4i)
(1, 1)
(0, −1)
1 − 6i
4 + 4i
−5 + 2i
−2 + 1i
2 + 1i
−2i
(33, 17 + 30i)
(12, 7 + 12i)
(1, 1)
(4, 1 + 4i)
(1, 1 + 2i)
6 + 3i
−5 + 3i
−1 − 6i
3 + 2i
−3i
−3 + 1i
(36, 1 + 12i)
(25, 1 + 10i)
(1, 1)
(0, −1)
(16, 1 + 8i)
6 + 5i
−6 + 1i
−6i
1 + 1i
−1i
−1
(40, 31 + 40i)
(24, 17 + 24i)
(1, 1 + 2i)
(1, 1)
(12, 7 + 12i)
−8i
5 + 4i
−5 + 4i
−4 + 3i
4 + 1i
−4i
(49, 1 + 14i)
(36, 1 + 12i)
(1, 1)
(0, −1)
(25, 1 + 10i)
7 + 6i
−7 + 1i
−7i
1 + 1i
−1i
−1
(49, 1 + 28i)
(16, 1 + 8i)
(0, −1)
(9, 1 + 6i)
(1, 1)
−7 + 4i
−7i
7 + 3i
1 − 2i
−2
1 + 2i
(49, 1 + 42i)
(25, 1 + 20i)
(0, −1)
(4, 1 + 4i)
(9, 1 + 6i)
−7 + 5i
−7i
7 + 2i
2 − 3i
−3
1 + 3i
(52, 7 + 28i)
(16, 1 + 8i)
(9, 1 + 6i)
(1, 1)
(0, −1)
2 − 8i
5 + 6i
−7 + 2i
−2 + 1i
2 + 1i
−2i
(57, 17 + 42i)
(28, 7 + 20i)
(4, 1 + 4i)
(1, 1)
(9, 1 + 6i)
2 − 9i
5 + 6i
−7 + 3i
−3 + 2i
3 + 1i
−3i
INTEGER SUPERHARMONIC MATRICES
(60, 49 + 60i)
(40, 31 + 40i)
(1, 1 + 2i)
(1, 1)
(24, 17 + 24i)
−10i
6 + 5i
−6 + 5i
−5 + 4i
5 + 1i
−5i
(64, 1 + 16i)
(49, 1 + 14i)
(1, 1)
(0, −1)
(36, 1 + 12i)
8 + 7i
−8 + 1i
−8i
1 + 1i
−1i
−1
(64, 1 + 48i)
(25, 1 + 20i)
(9, 1 + 6i)
(0, −1)
(4, 1 + 4i)
8 + 5i
−8 + 3i
−8i
2 + 3i
1 − 3i
−3
(64, 31 + 56i)
(33, 17 + 30i)
(1, 1)
(4, 1 + 4i)
(12, 7 + 12i)
9 + 4i
−7 + 4i
−2 − 8i
4 + 3i
−4i
−4 + 1i
(72, 17 + 48i)
(28, 7 + 20i)
(1, 1)
(9, 1 + 6i)
(4, 1 + 4i)
8 + 6i
−8 + 3i
−9i
3 + 2i
−3i
−3 + 1i
(73, 49 + 70i)
(24, 17 + 24i)
(1, 1)
(12, 7 + 12i)
(1, 1 + 2i)
9 + 4i
−7 + 5i
−2 − 9i
5 + 3i
−5i
−5 + 2i
(76, 7 + 60i)
(25, 1 + 20i)
(4, 1 + 4i)
(9, 1 + 6i)
(0, −1)
−1 − 10i
8 + 4i
−7 + 6i
−4
2 + 3i
2 − 3i
(81, 1 + 18i)
(64, 1 + 16i)
(1, 1)
(0, −1)
(49, 1 + 14i)
9 + 8i
−9 + 1i
−9i
1 + 1i
−1i
−1
(81, 1 + 36i)
(25, 1 + 10i)
(0, −1)
(16, 1 + 8i)
(1, 1)
−9 + 5i
−9i
9 + 4i
1 − 2i
−2
1 + 2i
47
48
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
(81, 17 + 60i)
(28, 7 + 20i)
(9, 1 + 6i)
(4, 1 + 4i)
(1, 1)
−10 + 3i
3 − 9i
7 + 6i
−4i
−3 + 2i
3 + 2i
(81, 1 + 72i)
(49, 1 + 42i)
(0, −1)
(4, 1 + 4i)
(25, 1 + 20i)
−9 + 7i
−9i
9 + 2i
3 − 4i
−4
1 + 4i
(84, 7 + 36i)
(25, 1 + 10i)
(16, 1 + 8i)
(1, 1)
(0, −1)
3 − 10i
6 + 8i
−9 + 2i
−2 + 1i
2 + 1i
−2i
(84, 71 + 84i)
(60, 49 + 60i)
(1, 1 + 2i)
(1, 1)
(40, 31 + 40i)
−12i
7 + 6i
−7 + 6i
−6 + 5i
6 + 1i
−6i
(88, 49 + 80i)
(33, 17 + 30i)
(12, 7 + 12i)
(1, 1)
(4, 1 + 4i)
−11i
8 + 6i
−8 + 5i
−5 + 3i
5 + 2i
−5i
(96, 31 + 72i)
(57, 17 + 42i)
(4, 1 + 4i)
(1, 1)
(28, 7 + 20i)
3 − 12i
6 + 8i
−9 + 4i
−4 + 3i
4 + 1i
−4i
(97, 49 + 92i)
(33, 17 + 30i)
(4, 1 + 4i)
(12, 7 + 12i)
(1, 1)
−9 + 7i
−1 − 10i
10 + 3i
1 − 6i
−5 + 2i
4 + 4i
(100, 1 + 20i)
(81, 1 + 18i)
(1, 1)
(0, −1)
(64, 1 + 16i)
10 + 9i
−10 + 1i
−10i
1 + 1i
−1i
−1
(100, 1 + 60i)
(49, 1 + 28i)
(0, −1)
(9, 1 + 6i)
(16, 1 + 8i)
−10 + 7i
−10i
10 + 3i
2 − 3i
−3
1 + 3i
INTEGER SUPERHARMONIC MATRICES
(105, 49 + 90i)
(64, 31 + 56i)
(1, 1)
(4, 1 + 4i)
(33, 17 + 30i)
12 + 5i
−9 + 5i
−3 − 10i
5 + 4i
−5i
−5 + 1i
(108, 17 + 60i)
(52, 7 + 28i)
(9, 1 + 6i)
(1, 1)
(16, 1 + 8i)
4 − 12i
6 + 9i
−10 + 3i
−3 + 2i
3 + 1i
−3i
(112, 97 + 112i)
(84, 71 + 84i)
(1, 1 + 2i)
(1, 1)
(60, 49 + 60i)
−14i
8 + 7i
−8 + 7i
−7 + 6i
7 + 1i
−7i
(121, 1 + 22i)
(100, 1 + 20i)
(1, 1)
(0, −1)
(81, 1 + 18i)
11 + 10i
−11 + 1i
−11i
1 + 1i
−1i
−1
(121, 1 + 44i)
(36, 1 + 12i)
(0, −1)
(25, 1 + 10i)
(1, 1)
−11 + 6i
−11i
11 + 5i
1 − 2i
−2
1 + 2i
(121, 1 + 66i)
(49, 1 + 28i)
(16, 1 + 8i)
(0, −1)
(9, 1 + 6i)
11 + 7i
−11 + 4i
−11i
2 + 3i
1 − 3i
−3
(121, 1 + 88i)
(64, 1 + 48i)
(9, 1 + 6i)
(0, −1)
(25, 1 + 20i)
11 + 8i
−11 + 3i
−11i
3 + 4i
1 − 4i
−4
(121, 1 + 110i)
(81, 1 + 72i)
(0, −1)
(4, 1 + 4i)
(49, 1 + 42i)
−11 + 9i
−11i
11 + 2i
4 − 5i
−5
1 + 5i
49
50
LIONEL LEVINE, WESLEY PEGDEN, AND CHARLES K. SMART
(124, 7 + 44i)
(36, 1 + 12i)
(25, 1 + 10i)
(1, 1)
(0, −1)
4 − 12i
7 + 10i
−11 + 2i
−2 + 1i
2 + 1i
−2i
(129, 17 + 66i)
(52, 7 + 28i)
(1, 1)
(16, 1 + 8i)
(9, 1 + 6i)
10 + 9i
−11 + 3i
1 − 12i
3 + 2i
−3i
−3 + 1i
(129, 97 + 126i)
(40, 31 + 40i)
(1, 1)
(24, 17 + 24i)
(1, 1 + 2i)
12 + 5i
−9 + 7i
−3 − 12i
7 + 4i
−7i
−7 + 3i
(136, 31 + 88i)
(72, 17 + 48i)
(1, 1)
(9, 1 + 6i)
(28, 7 + 20i)
12 + 8i
−11 + 4i
−1 − 12i
4 + 3i
−4i
−4 + 1i
(144, 1 + 24i)
(121, 1 + 22i)
(1, 1)
(0, −1)
(100, 1 + 20i)
12 + 11i
−12 + 1i
−12i
1 + 1i
−1i
−1
(144, 1 + 120i)
(49, 1 + 42i)
(25, 1 + 20i)
(0, −1)
(4, 1 + 4i)
12 + 7i
−12 + 5i
−12i
3 + 5i
2 − 5i
−5
(144, 127 + 144i)
(112, 97 + 112i)
(1, 1 + 2i)
(1, 1)
(84, 71 + 84i)
−16i
9 + 8i
−9 + 8i
−8 + 7i
8 + 1i
−8i
INTEGER SUPERHARMONIC MATRICES
(145, 49 + 110i)
(96, 31 + 72i)
(4, 1 + 4i)
(1, 1)
(57, 17 + 42i)
4 − 15i
7 + 10i
−11 + 5i
−5 + 4i
5 + 1i
−5i
(148, 7 + 84i)
(49, 1 + 28i)
(9, 1 + 6i)
(16, 1 + 8i)
(0, −1)
−1 − 14i
11 + 6i
−10 + 8i
−4
2 + 3i
2 − 3i
(148, 97 + 140i)
(73, 49 + 70i)
(1, 1)
(12, 7 + 12i)
(24, 17 + 24i)
14 + 5i
−10 + 7i
−4 − 12i
7 + 5i
−7i
−7 + 2i
(153, 17 + 84i)
(52, 7 + 28i)
(16, 1 + 8i)
(9, 1 + 6i)
(1, 1)
−14 + 3i
5 − 12i
9 + 9i
−4i
−3 + 2i
3 + 2i
(153, 17 + 120i)
(76, 7 + 60i)
(4, 1 + 4i)
(9, 1 + 6i)
(25, 1 + 20i)
−2 − 15i
11 + 6i
−9 + 9i
−6
3 + 4i
3 − 4i
(156, 7 + 132i)
(49, 1 + 42i)
(4, 1 + 4i)
(25, 1 + 20i)
(0, −1)
−3 − 14i
12 + 4i
−9 + 10i
−6 − 1i
2 + 5i
4 − 4i
(156, 71 + 132i)
(105, 49 + 90i)
(1, 1)
(4, 1 + 4i)
(64, 31 + 56i)
15 + 6i
−11 + 6i
−4 − 12i
6 + 5i
−6i
−6 + 1i
51